This publication like any my books does precisely what the identify says other than there are numerous greater than 10 rules given during this e-book. I in simple terms aspect out 10 principles yet there are many extra given. this is often written for these desirous about what they're going to do after they get right here and get up on a daily basis. Retirement could be a drag in case you do not have enjoyable activities and particularly if you happen to wouldn't have sufficient funds to do the belongings you are looking to do. So if you happen to imagine it's possible you'll wish to upload for your source of revenue and even as, aid another humans might be, and hold your self busy all while - this booklet is simply what you must seriously look into. Doing company might be difficult when you do not know the best way to do it legally in a brand new position. This publication will stroll you thru a number of principles that i do know will give you the results you want that are effortless and fun.
I have written eight books approximately the Philippines; 2 extra holiday or brief stopover at orientated, and six books every one with a distinct concentration. while you are fascinated about relocating to the Philippines then I recommend you begin with the 1st ebook I wrote at the Philippines which is simply entitled THE PHILIPPINES. If you haven't learn that e-book, you're not getting the entire tale by means of simply choosing and selecting of the opposite books. They are broke up into diversified books instead of 1 tremendous e-book simply because i do not believe too many of us will be clicking on purchase now if the booklet used to be six hundred pages and was once priced at $69.95.
The method i've got performed this for you is so that you can learn a hundred pages or so of excellent details and while you're no longer scared off by way of that info, then you definately should buy one other of my books with a purpose to offer you additional details in one other course etc etc. By the time a significant one who is basically eager to recognize if relocating to the Philippines is true for them has learn the 6 books i've got written at the topic, they're going to have good enough info to grasp if that circulation is correct for them or not.
Once you are making the choice to maneuver right here, retired right here, as i've got, then you definitely can search for extra precise details that is on hand unfastened on the web approximately varied Visa's which are on hand and so forth and a few of the extra legalistic details. That stuff is altering for all time and so it is difficult to place that during a ebook or even a 12 months later have it nonetheless be stable present details. Google is your most sensible wager for that sort of fluid details. What my books will inform you is what i think you want to learn about so that you could make a decision even if retirement within the Philippines is correct for you.
Takuan Sōho's (1573–1645) works on Zen and swordsmanship are one of the simplest and energetic displays of Zen ever written and feature loved nice recognition in either premodern and smooth Japan. even if dealing ostensibly with the artwork of the sword, checklist of Immovable knowledge and at the Sword Taie are easy publications to Zen—"user's manuals" for Zen brain that express one the right way to show up it not just in sword play yet from second to second in daily life.
India moves its customer with a sensory, highbrow, non secular and philosophical assualt that is unrivaled via the other position in the world: this subcontinent's sheer variety is amazing. Footprint's nineteenth variation India guide will advisor you from tropical seashore paradises and luxurious backwaters, to forests the place elephants roam and foothills clad in tea plantations.


Bali was a great Asura king who had conquered the heavens. At Indra's behest, Vishnu was born as a brahmin boy in the womb of Aditi, and went to the place where Bali was performing a great sacrifice. This is the Vamana Avatar of Lord Vishnu (Vamana = Dwarf).
Bali, despite being warned by his Guru Shukra that this was Vishnu in disguise, announced that he would not swerve from his oath, that of never refusing a supplicant. The Vamana then asked for three steps of his feet. In order to complete the donation, Bali would have to pour water from a container with a spout. Shukra made one last ditch attempt, by transforming into an insect and blocked the spout. Vamana however, pushed the insect away with the help of a blade of grass. (This punctured one eye of Shukra, who was blind in one eye from this point onwards). The gift was duly sanctified. At this point, Vamana assumed the Vishwaroopa (literally Universal-form) and with his first two steps measured the earth and the heavens. He then asked Bali, "Where should I put my third step?". Bali humbly replied, "Place it on my head". Thereupon Vishnu resumed his Vamana form and granted salvation to Bali by placing his tiny foot on Bali's head.
After this, Vamana restored the rule of the Gods on the heaven and earth and sent Bali to be the overlord of the nether-world.


I'm not able to reply to tickets when using FF 3 beta 4. When I try IE7 instead, it works fine.
Maybe it's time to upgrade Trac? The Trac demo on the official site seems to work with FF 3, but I wasn't logged in there, so I guess it's not a reliable test.


This photograph of St Pauls Cathedral shows the building floodlit in red, alongside the Millennium Bridge at night. This print is available as an art print, canvas or acrylic panel in various different sizes. To see more photographs of this location or theme, or for help or advice please call 01992 535344.


Celuk is a village with a famous silver art production, thus became a tourist attraction identical as the famous silver producer in Bali. Not only produce the silver craft but also gold craft. As a famous village, Celuk became the destination of domestic and foreign tourists. Travelers can come and visit here to see how the artist working on traditional crafts especially how creative they are to make gold jewelry with high quality production. You can easily find or can be purchased the silver craft.
Most people in Celuk is gold and silver craftsmen which creative, innovative also skilled in the art to develop silver and gold craft with creative variations style and design. The Silver products of Celuk Village have entered the local, national, and penetrated into the international market. Many types of jewelry creations and variations, either as souvenirs or export commodities has been produced in this village such as rings, bracelets, necklaces, earrings, studs, Bross and other types of jewelry.
Celuk Village is located in Sukawati, Gianyar, Bali. Celuk is located in a strategic location, when you are traveling tour and heading to Kintamani or Ubud you will be pass this place and can be used as an alternative destination to visit if you love the art of silver handicrafts of Bali island. In addition to join the tour you can also rent a car in Bali and visiting interesting places around it such as: Batubulan, Batuan and Tohpati.


If the water in your home has a chlorine smell or you don't like the taste, home filtration methods are available to give you a more satisfying drink of water. A few filters remove illness-producing microorganisms. Some water filters remove heavy minerals to soften your water, too. Here is a list of water filters to clean up your water, from DIY units to professionally installed systems.
This is the most basic of water filters. It uses small granules of carbon. These act as a physical filter, removing large particles from the water, as well as a chemical filter to remove chlorine and some pesticides. You can purchase these filters as a simple water pitcher or as a device that connects to your faucet to filter the water when you turn it on.
Maintenance of these systems requires replacing the carbon granules periodically. Carbon filters are often used in combination with other water filtration methods.
In this systems, water flows through a ceramic sleeve. Ceramic is porous and, as the water moves through the tiny holes, parasites and some minerals are removed. These are purely physical filters but may be used with a carbon filter to remove more contaminants from the water. Because this filter removes minerals it has some water softening effect.
You can buy ceramic units that attach to the faucet or to the water supply line under the sink. The ceramic pores can become clogged, restricting the water flow, so the sleeves need to be scrubbed periodically to open up the pores.
These filters kill bacteria, viruses and parasites in the water. They consist of ultraviolet light tubes in a cylinder through which the water flows. As the water passes over the UV tubes, harmful microorganisms are killed. These units must be installed by a plumber to make sure the UV tubes are adjusted for peak performance.
These filters are often combined with other types of filtration to remove particles, chlorine, pesticides and other chemicals. The light tubes fade over time and need to be replaced to continue keeping your water free of harmful bacteria and viruses.
Zinc and copper granules combine to create ions that pull minerals, parasites, bacteria and chemicals from the water. These filters remove heavy minerals and soften the water. Carbon filters are often used with this filter to remove more contaminates from the water.
The copper and zinc granules need to be replace periodically to maintain the balance between these two minerals. A water filtration company must install these units in your home to make sure the minerals are working together properly.
Contact a service such as Garabedian Plumbing & Heating Inc to learn more.


/Beliefs - Non-Serving Vs. Serving.
The point of the above example is that when we are young, we find it possible and very easy to be anything we want to be. We are able to believe this with any real effort.
The other way of putting this - Take your life of a few years ago, let's say 5 years ago and for a minute ponder on your life and in particular, your beliefs, dreams and desires and compare it to your beliefs and reality of today. Then reflect on your beliefs, dreams and desires of 10 years ago and compare it to your beliefs and reality of today.
The simple point is that when we are younger we believe we can be and achieve whatever we want to be and whatever we want to achieve, however, years later when we take stock and look at the reality of our lives, something went wrong somewhere!
Whilst considering this, the first factor to consider is why is there such a discrepancy between the beliefs, dreams and desires we had when we were younger and our reality of years later! Secondly, we need to evaluate whether or not we are prepared to do "whatever it takes" to create a life of choice, live our desires, and live our dream.
Basically what happens is as the circle of life carries on, we get caught up in life and over and above this, there is the factor of beliefs that will always run our lives based on the beliefs, and in the end, the source we believe in will always create our reality.
Over the years we are exposed to millions of messages literally pouring into our brains and although we can't remember them, they certainly are there at a sub-conscious level.
These messages create beliefs, some of them serving and some of them non-serving. Also known as conditioning and programming.
The fact is all your desires, life of choice is within your reach providing you are prepared to take action and take steps that can change your life.
There's no getting away from the need to change the way you do things, the way you think about things, beliefs, etc, should you wish to change your life.
Until such time that you take the decision to change, you will keep on repeating/creating your previous results!
Work Your Wealth's range of courses have all been specifically engineered to achieve this so that you can live a life of choice!
We specialise in assisting people to break this vicious circle and trap that they find themselves in through highly interactive workshops and seminars.
You have a very simple choice, keep on procrastinating and thinking and dreaming about it or maybe even keep thinking and putting your hopes on winning the lotto or maybe even for your ship to come in, or maybe the right opportunity to present itself ….
You can take a firm decision, commit to it by taking action and get yourself on a course or workshop of your choice that will enable you to overcome whatever it is that's holding you back.
Your beliefs at a sub conscious level determine your outcome! We consistently produce certain outcomes in our lives….. if your outcome is not what you want it to be, it simply means there are certain beliefs that are holding you back, that is simply not serving you and where you want to go, nor what you want!
There is only one way to change the outcome in your life – change the cause – the belief!
The difficult part of course is to do it on your own! Our seminars have been designed specifically to help with this!
Our seminars not only teach you how, it helps you discover your limiting beliefs!
At the very same time the seminars will not only give you tools to change your beliefs, the seminar will help you change the belief there and then!


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Huts by the River, Comporta
Typologies Culture / Leisure
City Comporta
Country Portugal
Photographer Nelson Garrido
An old timber wharf stands the test of time in the Sado estuary nature reserve in Comporta, one hour south of Lisbon. Its identity remains beyond the material resistance of its construction. The project stems from the silent respect for this solid essence that allows to change or replace the wood pieces to rebuild two cabins without losing the primary values. The archetypal forms of the cabins, with hip roofs, evoke a primitive imaginary filled with simplicity and purity.
The project develops into two autonomous spaces: one to unwind with the support of a kitchen integrated in the same material of the walls; and the other as a sleeping area with a small bathroom and a shower, accessible from the inside and from the outside. The slope of the walls and ceilings adapts subtly to these functions, and generates different situations in each corner because neither of them follows a straight angle. This ambiguity of the details is transferred to a superior scale; because the two structures appear to be the same, but they aren't. Built entirely in recycled wood and exposed to the weather that will continue their change, the cabins merge and evolve with the laconic atmosphere, adapting to the seasons... [+]
Technical Info Mostrar Ocultar ficha técnica
Cabañas en el río Huts by the River
Silent Living
Aires Mateus / Francisco y Manuel Aires Mateus (socios partners); Maria Rebelo Pinto (jefa de proyecto project leader); Luz Jiménez, David Carceller (equipo team)
Contratista Contractor
Cenário Perfeito
Superficie construida Built-up area
Nelson Garrido
Aires Mateus Comporta Culture / Leisure Francisco Aires Mateus Manuel Aires Mateus Nelson Garrido Portugal Recycling Wood
House in an Old Winery, Azeitão
Aires Mateus Portugal The industrial buildings and old wine warehouses of the region of Azeitão (south of Lisbon) have gradually changed their function as the production methods have become more specialized. In Brejos, a small town just a few kilometers from the city, an
House in Melides I, Grândola
Aires Mateus Portugal A natural platform in the Melides mountains, just a few kilometers from the coast, provides the site for this holiday rental home, promoted by Primosfera: a Lisbon-based company which assesses the construction of properties with the same dedication a
Musée de l'Elysée and MUDAC, Lausanne
Aires Mateus Switzerland Promoted by the Canton of Vaud, the City of Lausanne, and the CFF (Swiss Federal Railways), the new Pôle Museal is an urban regeneration plan, of scope unprecedented in Switzerland, set out to transform the city into a dynamic center of cultural inno
Daytime Center, Grândola
Aires Mateus Portugal Located in Grândola, a small municipality in the district of Setúbal, this daytime civic center for the elderly is part of the social services network of the Santa Casa da Misericórdia: a Portuguese institution founded in the 15th century, and that p
House in Leiria
Aires Mateus Portugal Known for the famous medieval castle after which it is named, the municipality of Leiria spreads by the banks of the Lis River. This compact single-family house goes up on a hill on the outskirts of the city, on a site in a high position overlooking
Barreira Antunes House, Grândola
Aires Mateus Portugal In portuguese 'alentejo' means 'beyond the Tagus,' and it is the region that stretches from the south of the river to the Algarve. Located in the coastal area of this region and one hundred kilometers south of Lisbon is the municipality of Grândola,
Wastewater Treatment Plant, Lisbon
Aires Mateus Portugal Throughout its history, the Alcântara neighborhood in Lisbon, located by the Tagus River and at the foot of the iconic 25 de Abril Bridge, has undergone several transformations that have altered its natural landscape, with the proliferation of roads
House in Campo de Ourique, Lisbon
Aires Mateus Portugal Campo de Ourique lies on one of the seven hills of Lisbon. Thanks to its location and to the geological composition of its subsoil, it is one of the areas that best resisted the catastrophe caused by the earthquake and ensuing tsunami in 1755. A slow
School Center, Vila Nova da Barquinha
Aires Mateus Portugal This new school complex is located in a low-density neighborhood on the outskirts of a small town by the Tagus River. The shape of the volume is determined by two essential factors: the functional requirements of the program and the anonymous urban c
Houses in the Sand, Comporta
Aires Mateus Portugal This group of rental houses sits right on the beach in the seaside town of Comporta, in the Alentejo. The project is a response to very specific conditions. The starting point was to turn a group of old fishermen houses into a small hotel that, witho
ZAC Batignolles
Aires Mateus AAVP architecture France With a composition of series, encounters, and indentations that make the facade a dazzling urban altarpiece, the building rises fifty meters over the horizon of Paris.
Residence for Elderly People, Alcácer do Sal
Aires Mateus Portugal Located in the municipality of Alcácer do Sal, some 90 kilometers from Lisbon, this building gives an original answer to the situation of the elderly in Western societies. This is an even greater challenge these days because of the gradual aging of t
AV Monografías 225 (pages 88-91) Aires Mateus Open the digital publication

redpajama_set_name: RedPajamaCommonCrawl


Here in the United States, in the whole Northern Hemisphere actually, we have hit the peak gardening month. This is the time of the year when we are reaping the rewards of our Winter and Spring work. Some crops are peaking, some are done and ready to be removed to make room for planting your Fall garden.
Plant your broccoli ten weeks before your first frost date. Because you are planting the seedlings while the summer days are still hot, mulch them to help retain moisture and to keep the soil cool.
Brussels sprouts are great for your Fall garden. The flavors are actually enhanced when harvested in the cooler weather. Start seeds in June and set plants out this month.
Start your cabbage and cauliflower plants six to eight weeks before your first frost date and transplant them into the garden after they have a couple sets of leaves. Like broccoli, they will need protection from the heat and sun and they need rich, fertile soil.
Sow turnips in the early part of the month to follow an early crop that is done.
Mustard, radish, kohlrabi, spinach and lettuce can all be sown as close as four to six weeks before your first frost date. Just pay attention to the particular variety's requirements.
It is not too late in most areas to sow a crop of shorter seasoned beans to be enjoyed as snap or green beans this fall.
When planting seeds in the middle of summer, remember that you may need to plant a bit deeper than normal since the surface layer of soil will dry out quickly.
Harvesting and processing of beans, carrots and cabbage is done now while a bit on the young side.
Side dressing heavy feeding plants like squash can be done by working well composted steer manure or compost.
While watering, use the time to inspect plants and remove any unwanted insects or egg clusters.
Where tomatoes are subject to sun-scald, pick the fruit as they first start to turn color and let them finish ripening indoors. Click here for more information about tomato problems and disease.
Bachelor buttons can be helped to bloom a second time by removing about six inches from the tops of the plants and fed a manure tea. Delphiniums can also be encouraged in a similar way by first removing the old flower stalk.
Sowing fresh delphiniums and Hollyhock seeds now will give best germination results and provide nice plants for next year.
Planting columbine from seed can be done by pressing into the soil so that they are no more than 1/8″ deep. The seeds often take up to 4 weeks to germinate.
Transplant irises now by taking divisions from the newer or outside growth and immediately planting in the new location by covering the rhizomes with 1 to 2 inches of soil.
Cuttings of plants that will need to winter over in the house or greenhouse should be done now. Root clippings of geraniums, begonias, coleus, etc. in moist sand.
Roses should be thoroughly soaked once a week or as needed in very hot weather. Mulching is very beneficial.
Prune roses as soon as they are finished blooming.
Many biennial and perennial flower seeds do well if sown in the first part of the month.
Rock gardens require about as much watering during this time as any other part of the garden.
Lawns do better if allowed to remain at least 1-1/2″ tall. They will stay greener, resist weeds becoming established, and remain greener.
Remove suckers from any grafted shrubs or trees immediately as they appear.
Keep an eye out for marauding birds that can strip blueberry bushes, cherry trees and other berry plants. Mesh coverings, noisemakers, and reflective devices have varied levels of success.
The developing bunches of grapes can be protected by securing small paper sacks over them.
May is the month when most of us all head into our main gardening season. It is during this month that most vegetables can be planted.
Set out tomato plants when evening temperatures stay above 50 degrees F.
Melons should be planted as soon as all danger of frost is past. Plant extra seeds in the hill and cull as necessary.
Sow tender plants like lima beans, cucumbers, pumpkins, gourds, and squash after all danger of frost is past.
Broccoli and Brussels Sprouts can also be planted now.
Succession crops of radishes, lettuce and other greens can be planted for longer harvest times.
Every couple of days, loosen up the crust on cucumber, squash, and melons to help the seedlings to emerge and to help prevent dampening off.
Rhubarb plants can use a heavy dose of composted manure. spread around the plants and dig in well.
Eggplants and peppers can be set out later in the month in the North and earlier in the South.
Consider New Zealand Spinach as an alternative or adjunct to standard spinach. It does great in warmer climates and is a perennial in climates with milder winters.
If you are a beginning gardener, don't just plant bush beans. It is not much more difficult to cultivate pole-type or climbers and you will be rewarded with longer harvest periods and greater amounts for the same garden space. Check out our trellising ideas on our Bean Page.
Witloof chicory can be planted early this month. They will provide roots that can be used in the cellar for forcing in the Winter as a crop of 'Belgium' or 'White' Endive.
If frost threatens, cover tender plants with straw, buckets, or old blankets. Make sure not to crush the plants.
Don't forget to include kitchen herbs like thyme, parsley, sage, and mints.
Nearly all varieties of flower seeds may be sown in the garden at this time.
Plants received by mail should be allowed to soak in water for several minutes and planted as soon as possible.
Shasta Daisies, Forget-Me-Nots and other "clumping" plants can be divided.
Dahlias should be divided and planted now.
Plant annuals in between your waning tulip and daffodil bulbs.
Do not remove daffodil foliage until after it yellows and begins to brown.
Before your sweet peas fall over, make sure that you have provide some form of support.
Now is the time to train evergreen shrubs like arbor viatae, hemlock, spruce and yew into shapes.
Mulch blueberry plants heavily with sawdust. Douglas fir works excellent.
Mulching newly planted shrubs and trees with grass clippings, leaves, or peat to conserve moisture. Newly planted roses, perennials, and fruit trees will also benefit from this practice.


Beautiful 2 bedrromed villa with Pool to rent in Paphos Cyprus. The villa is fully air-conditioned and well equipped for a home-from-home feel.
Detailed description : Adonis Villa is a luxury, well-equipped, two bedroom property situated in Zeus Gardens, a small exclusive development of privately owned villas, ideally located in Kato Paphos. The villa overlooks the large shared pool, which is positioned directly outside the patio doors, making it particularly suitable for families with children, and is for the exclusive use of the residents and their guests. The villa is air conditioned, beautifully presented and equipped to a very high standard. The accommodation is perfect for families or couples. The property is south facing for all day sun and maximum sunshine during the winter months. The excellent location means that car hire is an option, not a necessity. Just a leisurely stroll away is the tourist area of Kato Paphos with its picturesque harbour, mediaeval fort, municipal beach, shops, restaurants and bars. Paphos International airport is approximately 20 minutes away and airport transfers can be arranged if necessary. This is a child-friendly, relaxing and comfortable property – ideal for a fun, family holiday or a quiet peaceful escape for discerning couples.
General description : The villa sleeps 6 plus a baby. Travel cot and high chair can be provided at no extra cost.
Kitchen area : The kitchen is fully equipped with cooker, microwave, washing machine and all the utensils and crockery you will need.
Living area : The living room is open plan and has patio doors leading out to the patio area.
Dining area : The open plan dining area has seating for 6.
Bathroom(s) : The bathroom has a power shower and full size bath. There is also a downstairs cloakroom.
Bedroom(s) : There are 2 bedrooms. The double room facing the pool at the back of the property has a king size bed. The twin room at the front of the property has 2 single beds and room for a further put-u-up bed if needed.
Outside : The villa has a balcony at the front and back off of each bedroom. There is a patio area at the back of the villa with table and chairs and sunbeds exclusively for the use of our guests.
Pricing notes : Prices range from £195 per week in low season to £375 per week in high season. The price includes all bills, air-con, linen& towels and cleaning.
Booking conditions : £100 per week deposit secures a booking. Balance payable 8 weeks before your arrival.


We are a family business established over 37 years ago.
We offer an incredible range of handcrafted furniture in solid woods such as oak, mahogany, blackwood, oregon, cypress, beech, saligna and yellow wood.
Our craftsmen use traditional methods to produce quality furniture to last for generations.
We specialise in marble topped bathroom vanities and accessories, plasma TV stands, kitchen islands, and a full range of office and household furniture. We also offer an extensive variety of restored and second hand country furniture.
We are situated at the Rotunda – which is a unique stopover with a variety of shops including a coffee shop.
From Johannesburg Take the offramp 96 - Turn right over the highway - 200m on the left hand side. From Durban Take the offramp 96 - Turn left - 200m on the left hand side.


Rugged - Steel spoke wheels deliver Rokon power and torque on any terrain.
Climb - Climb a 60% grade. Rokons have climbed mountains rom the Rockies of Colorado to the Andes of Chile. With a ground clearance of 13 inches, the Scout can go up and over any terrain.
Legendary Rokon power and torque in a basic, rugged design.


The Wyomoco Resident Camp Program addresses the developmental needs and interests of campers within the goals and objectives of 4-H Youth Development and Cornell Cooperative Extension of Wyoming County. The program intends to help campers' ages 8 through 16 grow and learn in a healthy, positive environment; give them a wonderful, fun-filled week at camp, and provide a safe place for both these things to happen. 4-H Camp Wyomoco is fully owned and operated by Cornell Cooperative Extension of Wyoming County.
4-H Camp Wyomoco is coeducational and open to any boy or girl who has reached their eighth birthday, but have not reached their 17th birthday by the start of their week at camp. It is not necessary to be a 4-H member. 4-H Camp Wyomoco provides equal program and employment opportunities. Each day at 4-H Camp Wyomoco is filled with a variety of programs and activities. These include camper, recreational, and cabin activities, and all camp events. Special events are planned on a weekly basis to fit the theme of the week. These events provide your child with opportunities to make new friends and develop lasting camp memories.
4-H Camp Wyomoco is located at 2780 Buffalo Road, Varysburg, NY 14167 (20 miles south of Batavia) and features 280 beautiful acres of woods, meadows, and a three-acre pond.
Registration is Sunday from 1:00PM to 3:00PM at the Main Lodge. The gate will open no earlier than 12:50PM. If you are going to be late, please call us at camp: (585) 535-7381. Your child's first meal at camp will begin at 5:30PM.
Camper dismissal is at 4:30PM on Friday. Detailed instructions will be available at registration.
Camp is closed over the weekend. Campers who are staying for another week must leave on Friday and come back to camp on Sunday.
Horse Campers (Week One Only): If you are bringing a horse to camp, you will receive specific instructions on arrival.
The following prices are for a week at Camp Wyomoco in 2018. If you have any questions about prices, feel free to contact us at: campwyomoco@cornell.edu or call 585-786-2251.
$10.00 Paid In Full discount applied to each camp week if paid in full by 4/15/19 (EXPIRED).
A non-refundable, non-transferable deposit of $50.00 must accompany your application for each week at camp. A separate application is required for each week of camp and for each camper. If using a Wyoming County 4-H Cookie Sale Campership, which reduces the balance to less than $50.00, pay the balance only.
Returned checks will be charged a $25 fee. This fee, in addition to the amount of the returned check, must then be paid in full by cash, money order or credit card.
The balance due for camp must be paid by the Friday preceding your camp session. Cornell Cooperative Extension of Wyoming County handles all registrations.
Deposits are non-refundable and non-transferable (a deposit cannot be applied to the balance of another week's camp fees, nor transferred to another camper's fee).
If a doctor's certificate is provided stating that the camper is unable to attend camp because of illness or injury, the camp fee, minus the deposit, will be refunded.
A letter from a principal or other school official is required for campers who cancel out of a week because they have to attend summer school. The balance, minus the deposit, will be refunded.
Refunds for the balance of the camp fee, less the deposit, are made ONLY for cancellations received in writing at least three weeks prior to the opening day of the camp session.
A camper who does not arrive, arrives late, leaves early, or attends only part of the session will not receive a refund.
If the camp nurse and the camp director decide that a camper should be sent home due to medical reasons, a refund for the balance of the camp period will be made on a prorated basis, minus deposit, computed as of noon each day.
If a camper is experiencing a difficult adjustment to camp or is sent home because of a behavior problem or a pre-existing condition, no refund will be granted.
Rustic but comfortable. All campers live in cabins with staff supervision. Flush toilets and showers are located in both the girl's village and boy's village for convenience. Our professional kitchen staff takes pride in preparing food that is both nutritious and appealing to children. There is plenty of wholesome food and all the milk you can drink! The first meal is Sunday supper and the last is Friday lunch.
The staff members that are selected to work at 4-H Camp Wyomoco are hired based on their experience and qualifications in working with children. They are dedicated and conscientious. All staff undergo a comprehensive reference and background check. Counselors live in cabins with campers and teach camp classes and activities.
A Camp Nurse is on staff at camp. Further medical care would be provided at Wyoming County Community Hospital in Warsaw. The camp is inspected at least three times a year and licensed by the N.Y.S. Department of Health as required by law. Reports concerning the camp are on file at the Wyoming County Health Department, 5362 Mungers Mill Road, Silver Springs, NY 14550. New York State Health Code requires that a completed and up-to-date CAMPER INFORMATION PACKET be submitted by at least three (3) weeks prior to attending camp. A doctor's signature is required for a camper to take any supplements, over the counter or prescription medications. These medications will be administered by the camp nurse. Campers who are sick will not be admitted to camp.
Campers are expected to comply with camp rules and regulations and conform to the daily routines of camp. Failure to do so may result in dismissal from camp without a refund for the remainder of that session. Campers and their parents should refer to the CAMPER CODE OF CONDUCT, located on page two of the Camper Information Packet, which describes in detail the guidelines for camper behavior.
Cell phones, lap tops, computers, tablets, DVD players, any electronic device that connects to the internet, lighters, any open flames, weapons of any kind including but not limited to knives, guns, BB guns, sling shops, etc., Alcohol, recreational drugs, tobacco products both smoked and chewed, E cigarettes of any kind including but not limited to vapes, JUUls etc.
Bringing any of the above items to camp will result in immediate dismissal from camp and loss of camp fees. 4-H Camp Wyomoco has a NO TOLERANCE policy regarding contraband items at camp.
Please label all items with camper's name.
Personal Property, including but not limited to musical instruments, fishing poles, and sports equipment, etc. are the camper's responsibility alone and camp is not liable for loss or damage.
The following is a suggested list of items to pack when coming to camp. Week One Horse Campers should also review the information on the horsemanship page.
Note: Everyone should have at least one pair of long pants for hiking. If you have them, bring two bathing suits and an extra pair of old shoes or sneakers. Be sure to bring an ample supply of socks. All campers are encouraged to wear socks!
Cell phones are not allowed at camp for any purpose.
Note: Try to bring unscented toilet articles. The bugs at camp really enjoy scented soaps and shampoos; so don't encourage them to bother you!


text:
\section{Introduction}

Denote by $M_{n}\{0,1\}$ the set of 0-1 matrices of order $n$.  In 2007, Zhan \cite{ZH} proposed the following problem.
\begin{problem}\label{p1}
Given two integers $n$ and $k$, what is the maximum number of nonzero entries in a matrix $A\in M_n\{0,1\}$ such that $A^k\in M_n\{0,1\}$, and what are the extremal matrices attaining this maximum number?
\end{problem}
Wu \cite{WU} solved the case  $k=2$. Huang and Zhan \cite{HZ1} solved the case $k\ge n-1$ and they attained the maximum number for the case $k=n-2$ and $n-3$.  The authors of \cite{HLQ} solved the case $k\ge 5$ and they attained the maximum number for the case $k=4$. In this paper, we consider the following related problem.
\begin{problem}\label{p2}
	Given   integers $n$ and $k$, determine the maximum number of nonzero entries in a matrix $A \in M_n\{0,1\}$  such that $$tr(A)=0 {\rm ~and~}  A^k\in M_{n}\{0,1\}.$$ Characterize the extremal matrices that attain this maximum number.
\end{problem}
We solve the case $k=2$ for Problem \ref{p2} in this paper. Our  approach  is transferring the problem to an equivalent problem on digraphs and applying detail analysis on the structures of certain digraphs.  We need the following definitions and notations.

 We abbreviate directed
	walks, directed paths and directed cycles as walks, paths and cycles, respectively. The {\it length} of a walk, path or cycle is its number of arcs. The number of vertices in a digraph is called its {\it order} and the number of arcs its {\it size}. Let $u,w$ be two vertices. The notation $(u,w)$ or $u\rightarrow w$ means there exists an arc from $u$ to $w$;  $u\nrightarrow w$ means there exists no arc from $u$ to $w$;  $u\leftrightarrow w$ means both $u \rightarrow w$ and $w\rightarrow u$. If there exists an arc from $u$ to $w$, then $u$ is a {\it predecessor} of $w$, and $w$ is a {\it successor} of $u$.

 Let $D=(\mathcal{V},\mathcal{A})$ be a digraph with vertex set $\mathcal{V}=\{v_1,v_2,\ldots,v_n\}$ and arc set $\mathcal{A}$. Its {\it adjacency matrix} $A_D=(a_{ij})$ is defined by
\begin{equation}\label{eqh1}
a_{ij}=\left\{\begin{array}{ll}
                 1,& (v_i,v_j)\in \mathcal{A};\\
                 0,&\textrm{otherwise}.\end{array}\right.
                 \end{equation}
Conversely, given an $n\times n$ 0-1 matrix $A=(a_{ij})$, we can define its digraph $D(A)=(\mathcal{V},\mathcal{A})$ on vertices $v_1,v_2,\ldots,v_n$ by (\ref{eqh1}), whose adjacency matrix is $A$.

For a subset $X\subset \mathcal{V}$, $D(X)$ denotes the subdigraph of $D$ induced by $X$. The outdegree $d^+(u)$ is the number of arcs with tail $u$ and the indegree $d^-(u)$ is the number of arcs with head $u$. For convenience, if $X=\{x\}$ is a singleton, it will be abbreviated as $x$.

 For $S,T\subset \mathcal{V}$, denote by $\mathcal{A}(S,T)$ the set of arcs from $S$ to $T$. If $S=T$ we use $\mathcal{A}(S)$ instead of $\mathcal{A}(S,S)$. Let $e(S,T)=|\mathcal{A}(S,T)|$ and $e(D)=|\mathcal{A}(\mathcal{V})|$. $S\rightarrow T$ means for every vertex $i\in T$ there exists exactly one vertex $j\in S$ such that $j\rightarrow i$ ; $S\nrightarrow T$ means there is no arc from $S$ to $T$. If every vertex in $S$ has exactly one successor in $T$ and each vertex in $T$ has exactly one predecessor in $S$, we say $S$ {\it matches} $T$. Note that $S$ matching $T$ indicates $|S|=|T|$.

 For $W\subset \mathcal{V}$, denote by
 $$N^+_{W}(u)=\{x\in W|(u,x)\in \mathcal{A}\},$$
  $$N^-_{W}(u)=\{x\in W|(x,u)\in \mathcal{A}\},$$
  and $$N^+_{W}(S)=\bigcup\limits_{u\in S}N^+_{W}(u),$$ where $S\subset \mathcal{V}$. When $W=\mathcal{V}$, we simply write $N^+(u)$, $N^-(u)$ and $N^+(S)$ respectively.

The following problem is equivalent with Problem {\ref{p1} and Problem {\ref{p2}.
\begin{problem}\label{p3}
  Given two integers $n$ and $k$, determine the maximum size of  digraphs of order $n$ avoiding two distinct directed walks of a given length $k$ with the same initial and terminal vertices. Characterize the extremal digraphs attaining this maximum size.
\end{problem}
\noindent Using the digraphs of 0-1 matrices, we see that for a  matrix $A\in M_n\{0,1 \}$, $A^k\in M_n\{0,1\}$ if and only if $D(A)$ avoids distinct directed walks of length $k$ with the same initial vertex and the same terminal vertex.   Hence, for digraphs allow loops but donot allow parallel arcs,  Problem \ref{p3}  is equivalent with Problem \ref{p1}; for strict digraphs, i.e., digraphs do not allow loops or parallel arcs, Problem \ref{p3} is equivalent with Problem \ref{p2}.

For strict digraphs, the solution to Problem \ref{p3} for the case $k\ge 5$ follows straightforward from \cite{HLQ,HZ1},  since the extremal digraphs in \cite{HLQ,HZ1} are loopless.
In this paper we consider the case $k=2$ for Problem \ref{p3} on strict digraphs.

In what follows digraphs are strict. Given a family of digraphs $\mathscr{H}$ and a digraph $D$, $D$ is $\mathscr{H}$-free if $D$ contains no member of $\mathscr{H}$ as its subgraph. Let $\mathscr{F}=\{P_{2,2},C_{2,2}\}$ where $P_{2,2}$ and $C_{2,2}$ are defined as follows.

 \begin{figure}[H]
        \centering
        \includegraphics[width=1.1in]{P_22.jpg}\hspace{0.8cm}
        \includegraphics[width=1.5in]{F_22.jpg}\hspace{0.8cm}\\

     $P_{2,2}$\hspace{4.1cm}$C_{2,2}$
 \end{figure} It is obvious that a digraph avoids two distinct 2-walks with the same initial and terminal vertices if and only if it is $\mathscr{F}$-free. Denote by $ex(n)$ and $EX(n)$ the maximum size of $\mathscr{F}$-free digraphs of order $n$ and the set of $\mathscr{F}$-free digraphs of order $n$ attaining the maximum size, respectively.

Note that Problem \ref{p3} is a {\rm Tur{\'a}n\,} type problem, which concerns the study of  extremal graphs  that avoid given subgraphs. {\rm Tur{\'a}n\,} problem is a hot topic in graph theory with a long history; see \cite{BB,BH,BS,MS,PT,PT2}.   In \cite{HL}  the authors studied a closely related {\rm Tur{\'a}n\,} problem. They determined the maximum size of $P_{2,2} $-free digraphs as well as  the extremal $P_{2,2} $-free digraphs.
 \section{Main results}

If a digraph $D$ is acyclic and there is a vertex $u$ such that there is a unique directed path from $u$ to any other vertex, then we say $D$ is  an {\it arborescence} with root $u$.  If the maximum length of these paths is at most $r$, then we say $D$ is  an {\it $r$-arborescence}. Moreover, if $D$ is a 1-arborescence, we also say $D$ is an {\it out-star}. Throughout this article, we assume each arborescence has more than one vertices.

We will use $S(x)$, $S_{x}(y)$, $T(x)$ and $T(x,y)$ to denote the following digraphs, whose orders will be clear from the context. Note that $S(x)$ is an out-star with root $x$;
$S_{x}(y)$ is the union of a 2-cycle $x\leftrightarrow y$ and an out-star with root $y$; $T(x)$ is a  2-arborescence with root $x$; $T(x,y)$ is the union of a  2-cycle $x\leftrightarrow y$ and two 2-arborescences with roots $x$ and $y$.
\begin{figure}[H]
        \centering
        \includegraphics[width=.7in]{S.jpg}\hspace{0.8cm}
        \includegraphics[width=1in]{Sxy.jpg}\hspace{0.8cm}
        \includegraphics[width=1in]{Tx.jpg}\hspace{0.8cm}
        \includegraphics[width=2in]{T.jpg}\hspace{0.8cm}\\

       $S(x)$\hspace{1.5cm}$S_{x}(y)$\hspace{3.3cm}$T(x)$\hspace{3.4cm}$T(x,y)$
 \end{figure}

Here we denote a single vertex by $C_1$. The digraph obtained by taking the union of digraphs $D$ and $H$ with disjoint vertex sets is the disjoint union, written $D+H$. In general, $mD$ is the digraph consisting of $m$ pairwise disjoint copies of $D$. Denote by $D'$ the reverse of $D$, which is obtained by reversing the directions of all arcs of $D$. The reverse of an out-star is called an {\it in-star}.

Now we present the following six classes of digraphs on $n$ vertices, where $n$ is even for $D_1$ and odd for the others. Each of these diagraphs has vertex partition $\mathcal{V}_1\cup \mathcal{V}_2$ with $|\mathcal{V}_1|=\lfloor\frac{n}{2}\rfloor+1$ and $|\mathcal{V}_2|=\lceil\frac{n}{2}\rceil -1$.

\begin{figure}[H]
		\centering
		\includegraphics[width=1.6in]{di1.png}\hspace{0.1cm}
		\includegraphics[width=1.6in]{di6.png}\hspace{0.1cm}
			\includegraphics[width=1.6in]{di2.png} \\
		$D_1$\hspace{4cm}$D_2$\hspace{4cm}$D_3$\\
		\vspace{0.3cm}
		\includegraphics[width=1.6in]{di3.jpg}\hspace{0.1cm}
		\includegraphics[width=1.6in]{di1.png}\hspace{0.1cm}
		\includegraphics[width=1.6in]{di5.png} \\
		$D_4$\hspace{4cm}$D_5$\hspace{4cm}$D_6$\\

\end{figure}

	In $D_1$,
	$D(\mathcal{V}_1)=T(y_1,y_2)$;
	$D(\mathcal{V}_2)$ is empty;
	$f$ means $$\mathcal{V}_1\backslash \{y_1,y_2\}~ {\rm matches} ~\mathcal{V}_2;$$
	$g$ means $$u\rightarrow \mathcal{V}_1{\rm ~for~ all~} u\in \mathcal{V}_2.$$

	In $D_2$,
	$D(\mathcal{V}_1)=S_y(x)$; $D(\mathcal{V}_2)$ is the disjoint union of $T'(w)$ and isolated vertices, where $T'(w)$ may vanish. $\mathcal{V}_2$ is partitioned as $\mathcal{V}_2=\mathcal{V}_3\cup \mathcal{V}_4$, where $\mathcal{V}_3$ consists of all the isolated vertices of $\mathcal{V}_2$ and $|\mathcal{V}_3|\ge 1$;  $f$ means
	$$\mathcal{V}_1\backslash \{y,x\}{\rm ~matches~}\mathcal{V}_2\backslash \{w\} ~{\rm and~}x\rightarrow w;$$ $g$ means
	$$u\rightarrow \mathcal{V}_1 {\rm ~for~ all~} u\in \mathcal{V}_3 ~{\rm and}~ u\rightarrow \mathcal{V}_1\backslash  \{x\}{\rm ~for~ all~} u\in\mathcal{V}_4.$$

	In $D_3$, $D(\mathcal{V}_1)=T(y)$; $D(\mathcal{V}_2)$ is empty;  $f$ means $$\mathcal{V}_1\backslash \{y\}~ {\rm matches} ~\mathcal{V}_2;$$
	$g$ means $$u\rightarrow \mathcal{V}_1{\rm ~for~ all~} u\in \mathcal{V}_2.$$

	In $D_4$, $D(\mathcal{V}_1)=S(x)$;  $D(\mathcal{V}_2)$ is the disjoint union of $T'(w)$, in-stars and isolated vertices, and each of them may vanish. $\mathcal{V}_2$ is partitioned as $\mathcal{V}_2=\mathcal{V}_3\cup \mathcal{V}_4$, where $\mathcal{V}_3$ consists of all isolated vertices of $\mathcal{V}_2$ and the roots of $T'(w)$ and the in-stars; $f$ means
	$$\mathcal{V}_1\backslash \{x,y\}~ {\rm matches} ~\mathcal{V}_2\backslash \{w\} ~{\rm and ~}x\rightarrow w,$$  where $y$ is an arbitrary vertex in $\mathcal{V}_1\backslash \{x\}$; $g$ means
	$$u\rightarrow \mathcal{V}_1 {\rm ~for~ all~} u\in \mathcal{V}_3 ~{\rm and}~ u\rightarrow \mathcal{V}_1\backslash  \{x\}{\rm ~for~ all~} u\in\mathcal{V}_4.$$

	In $D_5$,  $D(\mathcal{V}_1)=T(y_1,y_2)$;
	$D(\mathcal{V}_2)$ is empty;
	$f$ means $$\mathcal{V}_1\backslash \{y_1,y_2\}{\rm ~matches~} \mathcal{V}_2\backslash \{x\},$$ where $x$ is an arbitrary vertex in $\mathcal{V}_2$;
	$g$ means $$u\rightarrow \mathcal{V}_1{\rm ~for~ all~} u\in \mathcal{V}_2.$$

	In $D_6$,  $D(\mathcal{V}_1)=S_y(x)$;  $D(\mathcal{V}_2)$ is the disjoint union of $T'(w,z)$ and isolated vertices. $\mathcal{V}_2$ is partitioned as $\mathcal{V}_2=\mathcal{V}_3\cup \mathcal{V}_4$, where $\mathcal{V}_3$ consists of all the isolated vertices of $\mathcal{V}_2$ and $|\mathcal{V}_3|\ge 1$;
	$f$ means
	$$\mathcal{V}_1\backslash \{y,y',x\}{\rm ~matches~} \mathcal{V}_2\backslash \{z,w\}~{\rm and}~x\rightarrow w,$$where $y'$ is an arbitrary vertex in $\mathcal{V}_1\backslash \{y,x\}$;  $g$ means $$u\rightarrow \mathcal{V}_1 {\rm ~for~ all~} u\in \mathcal{V}_3 ~{\rm and}~ u\rightarrow \mathcal{V}_1\backslash  \{x\}{\rm ~for~ all~} u\in\mathcal{V}_4.$$

We say that digraph $D $ is an isomorphism of $ H$ if there exists a bijection $f$: $\mathcal{V}(D)\rightarrow\mathcal{V}(H)$ such that $(u,v)\in \mathcal{A}(D)$ if and only if $(f(u),f(v))\in \mathcal{A}(H)$. Now we post our main result as follows.

\begin{theorem}\label{th1}

Let D be a digraph on $n$ vertices with $n\ge 8$. Then $$ex(n)=\lfloor\frac{n^2+4n}{4}\rfloor-1.$$ Moreover, $D\in EX(n)$ if and only if
\begin{itemize}
			\item[(1)] $n$ is even, and $D$ or $D'$ is an isomorphism of $D_1$;
			\item[(2)] $n$ is odd, and $D$ or $D'$ is an isomorphism of $D_i$ with $i\in \{2,\ldots,6\}$.
\end{itemize}
\end{theorem}
Equivalently, we have the following solution to Problem \ref{p2} when $k=2$.

\noindent {\bf Theorem \ref{th1}*}.
{\it Let $n\ge 8$ be an integer. Suppose $A\in M_n\{0,1\}$ such that
	$$tr(A)=0~{\rm and ~} A^2\in M_n\{0,1\}.$$ Then $A$ has  at most $ex(n)$ nonzero entries, and  $A$ has $ex(n)$  nonzero entries if and only if $D(A)$ or its reverse is an isomorphism of $D_i$ with $i\in \{1,\ldots,6\}$.}

  \section{Lemmas}
In this section we give preparatory lemmas for the proof of the main theorem.
\begin{lemma}\label{le8}
Let $n\ge 3$ be a positive integer. Then $D_i$ is $\mathscr{F}$-free  for $i=1,2,\ldots,6$. Moreover, we obtain
\begin{equation}\label{1.1}
ex(n)\ge
\begin{cases}
\frac{n^2+4n-5}{4},&\ n~is~odd;\\
\frac{n^2+4n-4}{4},&\ n~is~even.
\end{cases}
\end{equation}
\end{lemma}
\begin{proof}Suppose $D\in \{D_1,D_2,\ldots,D_6\}$ contains $u_1\rightarrow u_2\rightarrow u_4$ and $u_1\rightarrow u_3\rightarrow u_4$ with $u_2\ne u_3$.

For $i\in \{1,\ldots,6\}$, every vertex of $D_i$ has at most one predecessor in $\mathcal{V}_1$. Hence $u_2$ and $u_3$ can not both belong to $\mathcal{V}_1$. On the other hand, since each vertex has at most one successor in $\mathcal{V}_2$, $u_2$ and $u_3$ can not both belong to $\mathcal{V}_2$. Without loss of generality, we suppose $u_2\in \mathcal{V}_1$ and $u_3\in \mathcal{V}_2$. We have the following 4 cases.

{\it Case 1.} $u_1,u_4\in \mathcal{V}_1$. In $D_1,D_3,D_5$, each $2$-walk in $D(\mathcal{V}_1)$ originates at a vertex who has no successors in $\mathcal{V}_2$. In $D_4$, there exists no 2-walks in $D(\mathcal{V}_1)$. In $D_2,D_6$, among the vertices in $\mathcal{V}_1$, only $x$ has successors in both $\mathcal{V}_1$ and $\mathcal{V}_2$. The only 2-walk in $D(\mathcal{V}_1)$ with initial vertex $x$ is $x\rightarrow y\rightarrow x$. But the unique successor of $x$ in $\mathcal{V}_2$, namely $w$, is not a predecessor of $x$.

{\it Case 2.} $u_1\in \mathcal{V}_1$ and $u_4\in \mathcal{V}_2$. In $D_1$, $D_3$ and $D_5$, $D(\mathcal{V}_2)$ contains no arcs, and hence there exists no $u_3\rightarrow u_4$. In $D_2$ and $D_4$, among the vertices of $\mathcal{V}_1$ only $x$ has successors in both $\mathcal{V}_1$ and $\mathcal{V}_2$. We know $x$ has a unique successor in $\mathcal{V}_2$, namely $w$, but $w$ has no successor in $\mathcal{V}_2$. Then there exists no 2-walks from $x$ via $\mathcal{V}_2$ to $ \mathcal{V}_2$. In $D_6$, among the vertices of $\mathcal{V}_1$ only $x$ has successors in both $\mathcal{V}_1$ and $\mathcal{V}_2$. We know $x$ has a unique successor in $\mathcal{V}_2$, say $w$, and $w$ has a unique successor $z\in \mathcal{V}_2$. But there exists no 2-walk from $x$ to $z$ via $\mathcal{V}_1$.

{\it Case 3.} $u_1\in \mathcal{V}_2$ and $u_4\in \mathcal{V}_1$. In $D_1$, $D_3$ and $D_5$, $D(\mathcal{V}_2)$ contains no arcs. Then there exists no $u_1\rightarrow u_3$. In $D_2$, $D_4$ and $D_6$, if there exists $u_1$ has a successor $u_3\in \mathcal{V}_2$, then we have $N^+_{\mathcal{V}_1}(u_1)\nrightarrow N^+_{\mathcal{V}_1}(u_3)$.

{\it Case 4.} $u_1,u_4\in \mathcal{V}_2$. In $D_1$, $D_3$, $D_5$, $D(\mathcal{V}_2)$ contains no 2-walks, then $D$ contains no $u_1\rightarrow u_3\rightarrow u_4$. In $D_2$ and $D_4$, all 2-walks in $D(\mathcal{V}_2)$ end at $w$. For all $u\in \mathcal{V}_2$ with a successor in $\mathcal{V}_2$, we have $N^+_{\mathcal{V}_1}(u)=\mathcal{V}_1\backslash \{x\}$. Combining with $\mathcal{V}_1\backslash \{x\}\nrightarrow w$, there exists no $u_1\rightarrow u_2\rightarrow u_4$. In $D_6$, all 2-walks in $D(\mathcal{V}_2)$ end at $w$ or $z$. For all $u\in \mathcal{V}_2$ with successors in $\mathcal{V}_2$, we have $N^+_{\mathcal{V}_1}(u)=\mathcal{V}_1\backslash \{x\}$. Combining with $\mathcal{V}_1\backslash \{x\}\nrightarrow \{w,z\}$, there exists no $u_1\rightarrow u_2\rightarrow u_4$.

In the above cases, none of $D_1,\ldots,D_6$ contains the required 2-walks $u_1\rightarrow u_2\rightarrow u_4$ and $u_1\rightarrow u_3\rightarrow u_4$, a contradiction. Therefore, $D_1,\ldots,D_6$ are $\mathscr{F}$-free. By direct computation, we obtain $e(D_1)=\frac{n^2+4n-4}{4}$ and $e(D_i)=\frac{n^2+4n-5}{4}$ for $i=2,\dots,6$. Hence, we obtain (\ref{1.1}).

\end{proof}
Let $D=(\mathcal{V},\mathcal{A})$ be a digraph. For a fixed vertex $v\in \mathcal{V}$, denote by
\begin{equation*}\mathcal{V}_1(v)=N^+(v) ~{\rm and}~ \mathcal{V}_2(v)=\mathcal{V}\backslash \mathcal{V}_1(v),
 \end{equation*}\begin{equation*}
\mathcal{V}_3(v)=\{u\in \mathcal{V}_2(v)|N^+(u)=\mathcal{V}_1(v)\}~ {\rm and} ~\mathcal{V}_4(v)=\mathcal{V}_2(v)\backslash \mathcal{V}_3(v).
 \end{equation*}
 The index $v$ will be omitted if no confusion from the context.

 The following lemma is obvious.

	\begin{lemma}\label{le2}
			Let $D=(\mathcal{V},\mathcal{A})$ be an $\mathscr{F}$-free digraph. Then
		\begin{itemize}
			\item[(i)]   two distinct successors of a vertex share no common successor;
			\item[(ii)]given any $v\in \mathcal{V}$,  $e(\mathcal{V}_1(v),u)\le 1$ for all $u\in \mathcal{V}$.
		\end{itemize}
	\end{lemma}

\begin{lemma}\label{le3}
Let $D=(\mathcal{V},\mathcal{A})$ be an $\mathscr{F}$-free digraph and $v\in \mathcal{V}$. If there exists a 2-walk $$t_1\rightarrow t_2\rightarrow t_3 ~{\rm in}~ D(\mathcal{V}_1),$$ and $$N^+(u)=\mathcal{V}_1 ~{\rm for ~all} ~u\in\mathcal{V}_2,$$ then $t_1$ has no successor in $\mathcal{V}_2$.
\end{lemma}
\begin{proof}Suppose $t_1$ has a successor $t_4\in \mathcal{V}_2$. Since $N^+(t_4)=\mathcal{V}_1$, we have $t_4\rightarrow t_3$. Hence we obtain two distinct 2-walks from $t_1$ to $t_3$, a contradiction with $D$ being $\mathscr{F}$-free.
\end{proof}
 Let $\Delta^+(D)$ and $\Delta^-(D)$ denote the maximum outdegree and indegree of $D$. If no confusion arises, we write  $\Delta^+$ and $\Delta^-$, respectively.

\begin{lemma}\label{le4}
 Let $D=(\mathcal{V},\mathcal{A})\in EX(n)$. If $n$ is odd, then $\Delta^+\in \{\frac{n- 1}{2},\frac{n+ 1}{2}\}$; if $n$ is even, then $\Delta^+\in \{\frac{n}{2}-1,\frac{n}{2},\frac{n}{2}+1\}$.
\end{lemma}
\begin{proof} Let $v\in\mathcal{V}$ such that $d^+(v)=\Delta^+$. We count the size of $D$ in the following way
\begin{eqnarray*}
e(D)=e(\mathcal{V}_1,\mathcal{V})+e(\mathcal{V}_2,\mathcal{V})\le n+(n-\Delta^+)\Delta^+= -(\Delta^+-\frac{n}{2})^2+\frac{n^2+4n}{4}.
\end{eqnarray*}
The inequality is derived from Lemma \ref{le2} and the definition of $\Delta^+$. From (\ref{1.1}), we get the lemma.
\end{proof}

Let $$\alpha=\max\limits_{\substack{v\in \mathcal{V} ,\\ d^+(v)=\Delta^+  }}\max\limits_{u\in \mathcal{V}}{e(u,\mathcal{V}_2(v))}.$$ We give an upper bound on $\alpha$ as follows.
\begin{lemma}\label{le5}
 Let $D=(\mathcal{V},\mathcal{A})\in EX(n)$ and $n\ge 8$. Then $\alpha\le 1$.
 \end{lemma}

 \begin{proof} Suppose $\alpha\ge 3$. Then there exist $v$ and $u$ such that $d^+(v)=\Delta^+$ and $u$ has 3 successors $u_1,u_2,u_3\in \mathcal{V}_2(v)$. By Lemma \ref{le2}, we have $\sum\limits_{i=1}^{3}d^+(u_i)\le n$ and

\begin{eqnarray*}
e(D)&=& \sum\limits_{w\in \mathcal{V}_2\backslash\{u_1,u_2,u_3\}}d^+(w)+\sum\limits_{i=1}^{3}d^+(u_i)+\sum\limits_{w\in \mathcal{V}}e(\mathcal{V}_1,w)\\
&\le& (n-\Delta^+-3)\Delta^++n+n\\
&=&-(\Delta^+-\frac{n-3}{2})^2+\frac{n^2+2n+9}{4}.
\end{eqnarray*}
From (\ref{1.1}), we obtain $e(D)<ex(n)$, a contradiction. Hence, we have $\alpha\le 2$.

Now suppose $\alpha=2$. Then there exist $v$ and $u$ such that $d^+(v)=\Delta^+$ and $u$ has two successors $u_1,u_2\in \mathcal{V}_2(v)$. We claim that one of $u_1,u_2$ has $\Delta^+$ successors. Otherwise, we have $d^+(u_1)+d^+(u_2)\le 2\Delta^+-2$. Then
 \begin{eqnarray*}
 e(D)&=&\sum\limits_{w\in \mathcal{V}_2\backslash\{u_1,u_2\}}d^+(w)+\sum\limits_{i=1}^{2}d^+(u_i)+\sum\limits_{w\in \mathcal{V}}e(\mathcal{V}_1,w)\\
 &\le& (n-\Delta^+-2)\Delta^++ 2\Delta^+-2+n\\
 &=& (n-\Delta^+)\Delta^+-2+n\\
 &\le&\frac{n^2+4n}{4}-2.
 \end{eqnarray*}
 From (\ref{1.1}) we obtain $e(D)<ex(n)$, a contradiction. Hence we could assume $d^+(u_1)=\Delta^+$.

 By Lemma \ref{le2}, $u_2$ shares no common successor with $u_1$, i.e.,$$e(u_2,\mathcal{V}_1(u_1))=0.$$ By the definition of $\alpha$, we have $$e(u_2,\mathcal{V}_2(u_1))\le \alpha.$$ It follows that $d^+(u_2)\le \alpha= 2$. Hence, by (\ref{1.1}) we obtain
 \begin{eqnarray*}
e(D)&=& \sum\limits_{w\in \mathcal{V}_2\backslash\{u_1,u_2\}}d^+(w)+\sum\limits_{i=1}^{2}d^+(u_i)+\sum\limits_{w\in \mathcal{V}}e(\mathcal{V}_1,w)\\
&\le& (n-\Delta^+-2)\Delta^++\Delta^++2+n\\
&=&-(\Delta^+-\frac{n-1}{2})^2+\frac{n^2+2n+9}{4}\\
&<&ex(n).
\end{eqnarray*}
Therefore, we have $\alpha\le 1$.
\end{proof}

Let $D=(\mathcal{V},\mathcal{A})$ be a digraph. Let $v\in \mathcal{V}$ with $d^+(v)=\Delta^+$ and $u\in \mathcal{V}_2$. If $e(u,\mathcal{V}_1)=\Delta^+-1$, we denote the unique vertex of $\mathcal{V}_1\backslash N^+_{\mathcal{V}_1}(u)$ by $u'$.

\begin{lemma}\label{le6}
Let $D=(\mathcal{V},\mathcal{A})\in EX(n)$ with $n\ge 8$ and let $v\in \mathcal{V}$ with $d^+(v)=\Delta^+$. If $u_1\rightarrow u_2$ is in $D(\mathcal{V}_2(v))$, then $$N^+_{\mathcal{V}_1(v)}(u_1)\nrightarrow N^+_{\mathcal{V}_1(v)}(u_2).$$ Moreover, if $\mathcal{V}_1(v)\rightarrow \mathcal{V}_1(v)$ and $d^+(u_1)=d^+(u_2)=\Delta^+$, then $$u_1'\rightarrow N^+_{\mathcal{V}_1(v)}(u_2),~ u_2\in \mathcal{V}_4(v) ~{\rm and} ~N^+_{\mathcal{V}_1(v)}(u_2)=\mathcal{V}_1(v)\backslash \{u_1'\}.$$
\end{lemma}
\begin{proof}  Suppose $N^+_{\mathcal{V}_1(v)}(u_1)\rightarrow N^+_{\mathcal{V}_1(v)}(u_2)$. Then there exists $u_3\in N^+_{\mathcal{V}_1(v)}(u_1)$ and $u_4\in N^+_{\mathcal{V}_1(v)}(u_2)$ such that $u_3\rightarrow u_4$. We have $u_1\rightarrow u_3\rightarrow u_4$ and $u_1\rightarrow u_2\rightarrow u_4$, a contradiction.

For the second part, since $u_1\rightarrow u_2$ is in $D(\mathcal{V}_2(v))$, by Lemma \ref{le5} we have $e(u_1,\mathcal{V}_2(v))= 1$. It follows that $$e(u_1,\mathcal{V}_1(v))=\Delta^+-1{\rm ~ and~} N^+_{\mathcal{V}_1(v)}(u_1)=\mathcal{V}_1\backslash \{u_1'\}.$$
Since $N^+_{\mathcal{V}_1(v)}(u_1)\nrightarrow N^+_{\mathcal{V}_1(v)}(u_2)$ and $\mathcal{V}_1(v)\rightarrow \mathcal{V}_1(v)$, we have $u_1'\rightarrow N^+_{\mathcal{V}_1(v)}(u_2)$. It follows that $u_1'\notin N^+(u_2)$ and $u_2\in \mathcal{V}_4(v)$. By Lemma \ref{le5}, we have $e(u_2,\mathcal{V}_2(v))\le 1$ and $e(u_2,\mathcal{V}_1(v))\ge \Delta^+-1$. It follows that $N^+_{\mathcal{V}_1(v)}(u_2)=\mathcal{V}_1\backslash \{u_1'\}$.
\end{proof}
\begin{lemma}\label{le9}Let $D=(\mathcal{V},\mathcal{A})\in EX(n)$ with $n\ge 8$ and let $v\in\mathcal{V}$ with $d^+(v)=\Delta^+$. If $\mathcal{V}_1\rightarrow \mathcal{V}$ and $$d^+(u)=\Delta^+~{\rm ~for ~all ~}u\in \mathcal{V}_2,$$ then $$N^+(u)=\mathcal{V}_1~{\rm ~for ~all ~}u\in \mathcal{V}_2.$$
\end{lemma}
\begin{proof} Suppose there exists an arc $(u_1,u_2)$ in $D(\mathcal{V}_2)$. By Lemma \ref{le6}, we have $u_1'\rightarrow \mathcal{V}_1\backslash \{u_1'\}$ and $u_2\in \mathcal{V}_4$. Hence $u_2$ has a successor $u_3\in \mathcal{V}_2$. Since $u_1\rightarrow u_2\rightarrow u_3$ and $u_1\rightarrow \mathcal{V}_1\backslash \{u_1'\}$, we get $\mathcal{V}_1\backslash \{u_1'\}\nrightarrow u_3$. Since $\mathcal{V}_1\rightarrow \mathcal{V}$, we obtain $u_1'\rightarrow u_3$. By Lemma \ref{le6}, we have $u_3\in \mathcal{V}_4$. Then it has a successor $u_4\in \mathcal{V}_2$, which implies $u_1'\rightarrow u_3\rightarrow u_4$. By Lemma \ref{le5}, $u_1'\nrightarrow u_4$. From $\mathcal{V}_1\rightarrow \mathcal{V}$ we have $u_1'\rightarrow \mathcal{V}_1\backslash \{u_1'\}\rightarrow u_4$, which contradicts $D\in EX(n)$.
\end{proof}

\begin{lemma}\label{le7}
Let $D=(\mathcal{V},\mathcal{A})\in EX(n)$ with $n\ge 8$. If $\Delta^+\ge \Delta^-$, then $\Delta^+=\lceil\frac{n+1}{2}\rceil$.
\end{lemma}

\begin{proof} Let $v\in \mathcal{V}$ such that $d^+(v)=\Delta^+$. Since $\Delta^+\ge \Delta^-$, then
\begin{equation}\label{equ1}
e(\mathcal{V},\mathcal{V}_1)=\sum\limits_{u\in \mathcal{V}_1}d^-(u)\le \Delta^+\Delta^-\le (\Delta^+)^2.
\end{equation}
Applying Lemma \ref{le5} we obtain
\begin{equation}\label{equ2}
e(\mathcal{V},\mathcal{V}_2)=\sum\limits_{u\in\mathcal{V}\backslash \{v\}}e(u,\mathcal{V}_2)\le n-1.
\end{equation} It follows that
\begin{equation}\label{equ3}
e(D)=e(\mathcal{V},\mathcal{V}_1)+e(\mathcal{V},\mathcal{V}_2)
\le(\Delta^+)^2+n-1.
\end{equation} From (\ref{1.1}) we obtain $e(D)<ex(n)$ when $\Delta^+<\frac{n}{2}$. Hence, $\Delta^+\ge \frac{n}{2}$. Combining with Lemma \ref{le4}, we see that $\Delta^+=\frac{n+1}{2}$ when $n$ is odd and $\Delta^+\in \{\frac{n}{2},\frac{n}{2}+1\}$ when $n$ is even.

Now we consider the case $n$ is even and $\Delta^+=\frac{n}{2}$. By (\ref{1.1}), (\ref{equ3}) and $D\in EX(n)$, we have $e(D)=\frac{n^2+4n-4}{4}$. Combining with (\ref{equ1}), (\ref{equ2}) and (\ref{equ3}), we obtain
\begin{equation}\label{eq1}
d^-(u)=\Delta^+~{\rm for~all}~u\in \mathcal{V}_1.
\end{equation}and
$$e(\mathcal{V},\mathcal{V}_1)=\frac{n^2}{4},~e(\mathcal{V},\mathcal{V}_2)=n-1.$$
By Lemma \ref{le5}, each vertex in $\mathcal{V}\backslash \{v\}$ has a unique successor in $\mathcal{V}_2$.

By Lemma \ref{le2}, we have $e(\mathcal{V}_1,\mathcal{V})\le n$. It follows from (\ref{1.1}) that
$$e(\mathcal{V}_2,\mathcal{V})=e(D)-e(\mathcal{V}_1,\mathcal{V})\ge (n-\Delta^+)\Delta^+-1,$$ which implies there exists at least $\Delta^+-1$ vertices in $\mathcal{V}_2$ with outdegree $\Delta^+$.

Let $u_1\in \mathcal{V}_2\backslash \{v\}$ such that $d^+(u_1)=\Delta^+$ and $u_1$ has a successor $u_2\in \mathcal{V}_2$. We assert that either $u_2\nrightarrow u_1'$ or $u_1'$ has no predecessor in $\mathcal{V}_1$. Otherwise we have $u_2\rightarrow u_1'$ and there exists $u_1^*\in \mathcal{V}_1\backslash \{u_1'\}$ such that $u_1^*\rightarrow u_1'$. Then there exist two distinct 2-walks $u_1\rightarrow u_2\rightarrow u_1'$ and $u_1\rightarrow u_1^*\rightarrow u_1'$, a contradiction.
By Lemma \ref{le2}, we obtain $e(\mathcal{V}_1,u_1')\le 1$. It follows that
 $$d^-(u_1')=e(\mathcal{V}_2\backslash \{u_1,u_2\},u_1')+e(\{u_1,u_2\},u_1')+e(\mathcal{V}_1,u_1')\le \Delta^+-1,$$ which contradicts (\ref{eq1}). Hence, we have $\Delta^+=\frac{n}{2}+1$.

\end{proof}

\section{Proof of Theorem \ref{th1}}
Now we are ready to present the proof of Theorem \ref{th1}.
\begin{proof} Let $D=(\mathcal{V},\mathcal{A})\in EX(n)$. Note that $D\in EX(n)$ if and only if $D'\in EX(n)$. Without loss of generality, we assume $\Delta^+\ge \Delta^-$. Let $v\in \mathcal{V}$ such that $d^+(v)=\Delta^+$.  Keep in mind that given $u\in \mathcal{V}_2$ with outdegree $\Delta^+$, $u\in \mathcal{V}_4$ if and only if $u$ has exactly one successor in $\mathcal{V}_2$.

 By Lemma \ref{le2}, we have $$e(\mathcal{V}_1)\le |\mathcal{V}_1|~{\rm and ~}e(\mathcal{V}_1,\mathcal{V}_2)\le |\mathcal{V}_2|.$$
It follows that \begin{eqnarray}\label{equ3.2}
e(D)= e(\mathcal{V}_1)+e(\mathcal{V}_1,\mathcal{V}_2)+e(\mathcal{V}_2,\mathcal{V})\le |\mathcal{V}_1|+|\mathcal{V}_2|+|\mathcal{V}_2|\Delta^+=n+(n-\Delta^+)\Delta^+.
\end{eqnarray}We distinguish two cases according to the parity of $n$.

(1) $n$ is even. By Lemma \ref{le7}, we have $\Delta^+=\frac{n}{2}+1$. Then from (\ref{equ3.2}) we obtain
\begin{eqnarray*}
e(D)\le \frac{n^2+4n-4}{4}.
\end{eqnarray*}
It follows from (\ref{1.1}) that
\begin{equation}\label{equ4}
e(D)=ex(n)=\frac{n^2+4n-4}{4}.
\end{equation}
Now (\ref{equ3.2}) and (\ref{equ4}) lead to
\begin{equation*}
d^+(u)=\Delta^+~{\rm for~ all}~ u\in \mathcal{V}_2
\end{equation*}
 and
 $$e(\mathcal{V}_1)=|\mathcal{V}_1|~{\rm and}~e(\mathcal{V}_1,\mathcal{V}_2)=|\mathcal{V}_2|,$$ which implies
 \begin{equation}\label{e1}
\mathcal{V}_1\rightarrow \mathcal{V}_1 ~{\rm and}~\mathcal{V}_1\rightarrow \mathcal{V}_2.
\end{equation}
By Lemma \ref{le9}, we have
\begin{equation}\label{e3}
N^+(u)=\mathcal{V}_1 ~{\rm for~ all}~ u\in \mathcal{V}_2.
\end{equation}
Since $\alpha\le 1$, there exist $\frac{n}{2}-1$ vertices in $\mathcal{V}_1$ with exactly one successor in $\mathcal{V}_2$, leaving two vertices $y_1$ and $y_2$ in $\mathcal{V}_1$ with no successors in $\mathcal{V}_2$. Moreover,
  \begin{equation}\label{e2}
\mathcal{V}_1\backslash \{y_1,y_2\} ~{\rm matches}~ \mathcal{V}_2.
\end{equation}
\indent If there exists $z\in \mathcal{V}_1\backslash\{y_1,y_2\}$ such that $z\rightarrow y_2$, we know $z$ has a predecessor in $\mathcal{V}_1$. By Lemma \ref{le3}, we have $y_1\rightarrow z$ or $y_2\rightarrow z$. If the latter one holds, we obtain $z\rightarrow y_2\rightarrow z$, a contradiction with Lemma \ref{le3}. If the former one holds, we consider the predecessor of $y_1$ in $\mathcal{V}_1$. By Lemma \ref{le3}, we have $y_2\rightarrow y_1$. Thus, we obtain $z\rightarrow y_2\rightarrow y_1$, a contradiction with Lemma \ref{le3}. Hence, we have $y_1\rightarrow y_2$. Similarly, $y_2\rightarrow y_1$.

By (\ref{e1}) every vertex $u\in \mathcal{V}_1\backslash \{y_1,y_2\}$ has a unique predecessor $u_1\in \mathcal{V}_1$. By Lemma \ref{le3}, either $u\in N^+(y_1)\cup N^+(y_2)$ or $u_1\in N^+(y_1)\cup N^+(y_2)$. By the first part of (\ref{e1}), we have $D(\mathcal{V}_1)=T(y_1,y_2)$. Combining with (\ref{e3}) and (\ref{e2}), $D$ is an isomorphism of $D_1$.

(2) For $n$ is odd. By Lemma \ref{le7}, we have $\Delta^+=\frac{n+1}{2}$. It follows from (\ref{equ3.2}) that
\begin{equation}\label{equ5}ex(n)\le \frac{n^2+4n-1}{4}.
 \end{equation}Suppose equality in (\ref{equ5}) holds. From (\ref{equ3.2}), we obtain $d^+(u)=\Delta^+$ for all $u\in \mathcal{V}_2$ and $\mathcal{V}_1\rightarrow \mathcal{V}$. By Lemma \ref{le9}, (\ref{e3}) holds. Since $|\mathcal{V}_1|=|\mathcal{V}_2|+1$ and $\alpha\le 1$, there exists $y\in \mathcal{V}_1$ with no successor in $\mathcal{V}_2$ and
\begin{equation}\label{e4}
\mathcal{V}_1\backslash \{y\} ~{\rm matches}~ \mathcal{V}_2.
\end{equation}

Since $\mathcal{V}_1\rightarrow \mathcal{V}_1$, there must exist a cycle whose length is larger than or equal to 2. It follows that there exists a walk $t_1\rightarrow t_2\rightarrow t_3$ with $t_1\ne y$. By (\ref{e4}) $t_1$ has a successor $t_4\in \mathcal{V}_2$, which contradicts Lemma \ref{le3}.

Therfore, $ex(n)\le \frac{n^2+4n-5}{4}$. From (\ref{1.1}), we obtain $$e(D)=ex(n)=\frac{n^2+4n-5}{4}.$$ By Lemma \ref{le2} we have $e(\mathcal{V}_1,\mathcal{V})\le n$, which implies that $e(\mathcal{V}_2,\mathcal{V})\ge \frac{n^2-5}{4}$. Hence, at least $\frac{n-3}{2}$ vertices of $\mathcal{V}_2$ have outdegrees $\Delta^+$ and $d^+(u)\ge \frac{n-1}{2}$ for all $u\in \mathcal{V}_2$. Since $\alpha\le 1$, we have $$e(u,\mathcal{V}_1)\ge \frac{n-3}{2}~{\rm for~ all}~ u\in \mathcal{V}_2.$$
We will use the following claim repeatedly.

 {\it{\bf Claim 1.} Let $u\in \mathcal{V}_1$ with $e(u,\mathcal{V}_1)\ge 3$. If it has a predecessor $u^*$, then $u^*$ has no successors in $\mathcal{V}_2$.
 }

Otherwise, $u^*$ has a successor $u_1\in \mathcal{V}_2$. Combining with $e(u_1,\mathcal{V}_1)\ge \frac{n-3}{2}$ and $e(u,\mathcal{V}_1)\ge 3$, $N^+_{\mathcal{V}_1}(u)\cap N^+_{\mathcal{V}_1}(u_1)$ is not empty, which contradicts Lemma \ref{le2}.

As shown in (\ref{equ3.2}), the size of $D$ is the sum of three parts: $e(\mathcal{V}_1)$, $e(\mathcal{V}_1,\mathcal{V}_2)$, $e(\mathcal{V}_2,\mathcal{V})$. We have proven that the maximum size of $D$ is less by 1 than the sum of the maximum numbers of these three parts. Hence, we can deduce that two of three parts achieve the maximum numbers and one of them misses one from the maximum number. We present these three maximum numbers as follows.
\begin{equation}\label{e13}
e(\mathcal{V}_1,\mathcal{V}_1)=\Delta^+,
\end{equation}
\begin{equation}\label{e14}
e(\mathcal{V}_1,\mathcal{V}_2)=n-\Delta^+,
\end{equation}
and
\begin{equation}\label{e15}
 d^+(u)=\Delta^+ {\rm ~for~ all~} u\in\mathcal{V}_2.
\end{equation}
Moreover, (\ref{e13}) is is equivalent to
\begin{equation}\label{e16}
\mathcal{V}_1\rightarrow \mathcal{V}_1;
\end{equation}
and (\ref{e14}) is is equivalent to
\begin{equation}\label{e17}
\mathcal{V}_1\rightarrow \mathcal{V}_2.
\end{equation}
According to the above analysis, we distinguish three cases:

{\it Case 1.} (\ref{e13}) and (\ref{e14}) holds. Then $\frac{n-3}{2}$ vertices of $\mathcal{V}_2$ have outdegrees $\frac{n+1}{2}$ and a vertex $w\in \mathcal{V}_2$ has outdegree $\frac{n-1}{2}$. By Lemma \ref{le5}, we have $\alpha\le 1$. Then there exist $|\mathcal{V}_2|$ vertices of $\mathcal{V}_1$ having exactly one successor in $\mathcal{V}_2$, leaving only one vertex $y\in \mathcal{V}_1$ having no successor in $\mathcal{V}_2$. It follows that
\begin{equation}\label{e22}
\mathcal{V}_1\backslash \{y\} {\rm ~matches~} \mathcal{V}_2.
\end{equation}

Now we distinguish three cases.

{\it Subcase 1.1.} $D(\mathcal{V}_2)$ contains no arc. Then
\begin{equation}\label{e23}
N^+(u)=\mathcal{V}_1~{\rm for~all}~  u\in \mathcal{V}_2\backslash \{w\}~{\rm and~} N^+(w)\subset \mathcal{V}_1.
 \end{equation}
\indent We assert that if there exists $t_1\rightarrow t_2\rightarrow t_3$ in $D(\mathcal{V}_1)$, then either $t_1=y$ or $t_1\rightarrow w$. Otherwise, there exists $t\in \mathcal{V}_2\backslash \{w\}$ such that $t_1\rightarrow t$. From (\ref{e23}), we obtain a $2$-walk $t_1\rightarrow t\rightarrow t_3$, a contradiction.

Since $\mathcal{V}_1\rightarrow \mathcal{V}_1$, $D(\mathcal{V}_1)$ must contain a cycle. From the above assertion, we obtain $D(\mathcal{V}_1)$ contains only one cycle $z\leftrightarrow y$, where $z\in \mathcal{V}_1$ is the predecessor of $w$. Combining with $z\rightarrow w\rightarrow \mathcal{V}_1\backslash \{w'\}$, we obtain $z\notin \mathcal{V}_1\backslash \{w'\}$. Otherwise we have $z\rightarrow w\rightarrow z$ and $z\rightarrow y\rightarrow z$, a contradiction. Thus we have $z=w'$, which implies $w'\rightarrow w$. Moreover, $y$ has only one successor in $\mathcal{V}_1$, i.e.,
 \begin{equation}\label{e25}
N^+_{\mathcal{V}_1}(y)=w'
\end{equation}
\indent We assert
\begin{equation}\label{e24}N^+_{\mathcal{V}_1}(w')= \mathcal{V}_1\backslash \{w'\}.
 \end{equation}Otherwise, there exists $t\in \mathcal{V}_1\backslash \{w'\}$ with $w'\nrightarrow t$. We know $t$ has a predecessor $u\in \mathcal{V}_1$. Since $2$-walks in $D(\mathcal{V}_1)$ have to emanate from $y$ or $w'$, from (\ref{e25}) we have $w'\rightarrow u\rightarrow t$. At the same time, we have $w'\rightarrow w\rightarrow t$ since $N^+(w)=\mathcal{V}_1\backslash \{w'\}$, a contradiction. Hence, we obtain $w'\rightarrow \mathcal{V}_1\backslash \{w'\}$. It follows that
 \begin{equation}\label{e26}
 D(\mathcal{V}_1)=S_{y}(w').
\end{equation}
\indent Since (\ref{e22}) and $w'\rightarrow w$, we have $\mathcal{V}_1\backslash \{y,w'\}$ matches $\mathcal{V}_2\backslash \{w\}$. Combining with (\ref{e23}), (\ref{e24}) and (\ref{e26}), $D$ is an isomorphism of $D_2$ with $T'(w)$ vanishing.

{\it Subcase 1.2.} There exists an arc in $D(\mathcal{V}_2\backslash \{w\})$. Suppose $u_1\rightarrow u_2$ with $u_1,u_2\in \mathcal{V}_2\backslash \{w\}$. By Lemma \ref{le6}, we obtain $u_1'=u_2'$ and $u_1'\rightarrow \mathcal{V}_1\backslash \{u_1'\}$. Since $\mathcal{V}_1\rightarrow \mathcal{V}_1$, $u_1'$ has a predecessor $t\in \mathcal{V}_1$. By Claim 1 we have $y\rightarrow u_1'$. It follows that $$D(\mathcal{V}_1)=S_{y}(u_1').$$ Moreover, $u_1'$ has a unique successor $s\in \mathcal{V}_2$.

If $s\in \mathcal{V}_3$, we have $u_1'\rightarrow s\rightarrow u_1'$, a contradiction with $u_1'\rightarrow y\rightarrow u_1'$. If $s$ has a successor $s_1\in \mathcal{V}_2$, we have $u_1'\rightarrow s\rightarrow s_1$. On the other hand, from (\ref{e22}) we obtain $u_1'\rightarrow \mathcal{V}_1\backslash \{u_1'\}\rightarrow s_1$, a contradiction. Hence, we have
\begin{equation}\label{e27}
s=w ~{\rm and}~ N^+(w)\subset \mathcal{V}_1.
 \end{equation}
\indent We assert that
 \begin{equation*}\label{e28}
u_1'=w'.
 \end{equation*} Otherwise, we have $w\rightarrow u_1'\rightarrow \mathcal{V}_1\backslash \{u_1'\}$. On the other hand, since $w\rightarrow \mathcal{V}_1\backslash \{w'\}$, then $\mathcal{V}_1\backslash \{w'\}\nrightarrow \mathcal{V}_1\backslash \{u_1'\}$. Combining with (\ref{e16}), we obtain $w'\rightarrow \mathcal{V}_1\backslash \{u_1'\}$. Recalling $u_1'\rightarrow \mathcal{V}_1\backslash \{u_1'\}$, we obtain a contradiction with Lemma \ref{le2}. Then we have
 \begin{equation}\label{e29}D(\mathcal{V}_1)=S_{y}(w').
\end{equation}
\indent We assert that
\begin{equation}\label{e30}N^+_{\mathcal{V}_1}(u)=\mathcal{V}_1\backslash \{w'\}~ {\rm for ~all}~ u\in \mathcal{V}_4.
\end{equation}Otherwise, there exists a vertex $u\in \mathcal{V}_4\backslash \{w\}$ such that $u\rightarrow w'$. By Claim 1, we have $u$ has no successor in $\mathcal{V}_2$, which contradicts $u\in \mathcal{V}_4\backslash \{w\}$.

We assert that if there exists a 2-walk $u_3\rightarrow u_4\rightarrow u_5$ in $D(\mathcal{V}_2)$, then $u_5=w$. Otherwise, $u_5\ne w$. We have $u_3\ne w$ from (\ref{e27}). By (\ref{e22}) and $u_1'\rightarrow w$, we obtain $u_3\rightarrow \mathcal{V}_1\backslash \{u_1'\}\rightarrow u_5$, a contradiction. Therefore, for any $(t_1,t_2)$ in $D(\mathcal{V}_2)$, we have either $t_2=w$ or $t_2\in N^-(w)$. It follows that $D(\mathcal{V}_2)=T'(w)+aC_1$ with $a\ge 1$. Combining with (\ref{e22}), (\ref{e27}), (\ref{e29}) and (\ref{e30}), $D$ is an isomorphism of $D_2$.

Notice that there exists $(u_1, u_2)$ in $D(\mathcal{V}_2\backslash \{w\})$. There must exist a 2-walk in $D(\mathcal{V}_2)$.  Thus, in this case, $T'(w)$ must contain a 2-walk.

{\it Subcase 1.3.} $D(\mathcal{V}_2\backslash \{w\})$ contains no arc but $D(\mathcal{V}_2)$ contains arcs. We distinguish two cases.

{\it Subcase 1.3.1.} $N^+(w)\subset \mathcal{V}_1$. Then $\mathcal{V}_4\backslash \{w\}$ is not empty and
\begin{equation*}\label{e31}u\rightarrow w ~{\rm for~ all}~ u\in \mathcal{V}_4\backslash \{w\},
 \end{equation*}which leads to
 \begin{equation}\label{e34}
D(\mathcal{V}_2)=S'(w)+aC_1,
\end{equation}
where $a\ge 1$. By Lemma \ref{le6}, for any $u\in \mathcal{V}_4\backslash \{w\}$ we have $\mathcal{V}_1\backslash \{u'\}\nrightarrow \mathcal{V}_1\backslash \{w'\}$. Since $\mathcal{V}_1\rightarrow \mathcal{V}_1$, we have $u'\rightarrow \mathcal{V}_1\backslash \{w'\}$, which implies that $u'\notin \mathcal{V}_1\backslash \{w'\}$. It follows that
 \begin{equation}\label{eq3}u'=w' ~{\rm for~all}~u\in \mathcal{V}_4.
  \end{equation}Moreover, $w'\rightarrow \mathcal{V}_1\backslash \{w'\}$. From (\ref{e16}) $w'$ has a predecessor in $\mathcal{V}_1$. By Claim 1, we have $y\rightarrow w'$. Hence, we obtain
 \begin{equation}\label{e32}D(\mathcal{V}_1)=S_{y}(w').
 \end{equation}
\indent Let $w_1$ be the successor of $w'$ in $\mathcal{V}_2$. If $w_1$ has a successor $w_2\in \mathcal{V}_2$. Then by (\ref{e22}) we have $w'\rightarrow w_1\rightarrow w_2$ and $w'\rightarrow \mathcal{V}_1\backslash \{w'\}\rightarrow w_2$, a contradiction. If $w_1\in \mathcal{V}_3$, then $w_1\rightarrow w'$. Hence, we obtain $w'\rightarrow w_1\rightarrow w'$ and $w'\rightarrow y\rightarrow w'$, a contradiction. Thus, we get $w_1=w$, i.e., $w$ is the successor of $w'$ in $\mathcal{V}_2$. It follows from (\ref{e22})that
\begin{equation}\label{e33}\mathcal{V}_1\backslash \{y,w'\} ~{\rm matches}~ \mathcal{V}_2\backslash \{w\} ~{\rm and}~ w'\rightarrow w.
\end{equation} Combining with (\ref{e34}), (\ref{eq3}), (\ref{e32}) and (\ref{e33}), $D$ is an isomorphism of $D_{2}$, where $T'(w)$ is $S'(w)$. Note that we could consider $S'(w)$ as a special case of $T'(w)$.

{\it Subcase 1.3.2.} $w$ has a successor $w_0\in \mathcal{V}_2$. Let $\{w_1,w_2\}=\mathcal{V}_1\backslash N^+_{\mathcal{V}_1}(w)$. Suppose $N^+(w_0)=\mathcal{V}_1$. By Lemma \ref{le6}, we have $N^+_{\mathcal{V}_1}(w)\nrightarrow \mathcal{V}_1$. From (\ref{e16}), we have \begin{equation}\label{eq4}\{w_1,w_2\}\rightarrow \mathcal{V}_1.
 \end{equation}Since $D$ is loopless, we get
 \begin{equation}\label{eq5}w_1\leftrightarrow w_2.
  \end{equation}By (\ref{e22}) at least one of $\{w_1,w_2\}$ has a successor in $\mathcal{V}_2$. Without loss of generality, we assume $w_1$ has a successor $w_3\in \mathcal{V}_2$. If $N^+(w_3)=\mathcal{V}_1$, we have $w_1\rightarrow w_2\rightarrow w_1$ and $w_1\rightarrow w_3\rightarrow w_1$, a contradiction. Thus, $w_3\in \mathcal{V}_4$.

If $w_3\ne w$, since $w_1\rightarrow w_2\rightarrow N^+_{\mathcal{V}_1}(w_2)$ and $w_1\rightarrow w_3\rightarrow \mathcal{V}_1\backslash \{w_3'\}$, we have $N^+_{\mathcal{V}_1}(w_2)=w_3'$. From (\ref{eq4}) and (\ref{eq5}), we obtain
\begin{equation}\label{e35}
w_1=w_3'~{\rm and}~ D(\mathcal{V}_1)=S_{w_2}(w_1).
 \end{equation}
Since $w_3\in \mathcal{V}_4\backslash \{w\}$, $w_3$ has a successor $w_4\in \mathcal{V}_2$. Hence, from (\ref{e22}) and (\ref{e35}) we have $w_1\rightarrow w_3\rightarrow w_4$ and $w_1\rightarrow \mathcal{V}_1\backslash \{w_1\}\rightarrow w_4$, a contradiction. Therefore, we obtain $w_3=w$, i.e., \begin{equation}\label{e36}
w_1\rightarrow w.
\end{equation}
\indent Since $w_1\rightarrow w\rightarrow \mathcal{V}_1\backslash \{w_1,w_2\}$ and $w_1\rightarrow w_2$, we obtain $w_2\nrightarrow \mathcal{V}_1\backslash \{w_1,w_2\}$. Combining with $\{w_1,w_2\}\rightarrow \mathcal{V}_1$ and $w_2\nrightarrow w_2$, we have $w_1\rightarrow \mathcal{V}_1\backslash \{w_1\}$. It follows that
\begin{equation}\label{e37}D(\mathcal{V}_1)=S_{w_2}(w_1).
 \end{equation}
By (\ref{e22}) and (\ref{e36}), we have $\mathcal{V}_1\backslash \{w_1\}\rightarrow w_0$, which implies $w_1\rightarrow \mathcal{V}_1\backslash \{w_1\}\rightarrow w_0$. On the other hand, we have $w_1\rightarrow w\rightarrow w_0$, a contradiction.

Hence, $N^+(w_0)\ne \mathcal{V}_1$ and $w_0$ has a successor in $\mathcal{V}_2$. If this vertex is not $w$, then $D(\mathcal{V}_2\backslash \{w\})$ contains an arc, which contradicts the given condition of Subcase 3.3. Hence, we have
\begin{equation}\label{e38}w_0\rightarrow w.
 \end{equation}
 By Lemma \ref{le6}, we have $\mathcal{V}_1\backslash \{w_0'\}\nrightarrow \mathcal{V}_1\backslash\{w_1,w_2\}$. Since $\mathcal{V}_1\rightarrow \mathcal{V}_1$, we have $w_0'\rightarrow \mathcal{V}_1\backslash\{w_1,w_2\}$. Since $D$ is loopless, we have $w_0'\in \{w_1,w_2\}$. Without loss of generality, we suppose $w_0'=w_1$. It follows that
 \begin{equation}\label{eq6}w_0\rightarrow \mathcal{V}_1\backslash \{w_1\}.
 \end{equation} and
 \begin{equation}\label{e39}
 w_1\rightarrow \mathcal{V}_1\backslash\{w_1,w_2\}.
  \end{equation}By $w_0\rightarrow w\rightarrow w_0$ and(\ref{eq6}), we have $\mathcal{V}_1\backslash \{w_1\}\nrightarrow w_0$. Combining with (\ref{e22}), we obtain $$w_1\rightarrow w_0.$$ Similarly, since $w\rightarrow w_0\rightarrow w$ and $w\rightarrow \mathcal{V}_1\backslash \{w_1,w_2\}$, we have $\{w_1,w_2\}\rightarrow w.$ Hence, \begin{equation*}w_2\rightarrow w.\end{equation*}
\indent Applying Lemma \ref{le6} on $w\rightarrow w_0$, we have$$\mathcal{V}_1\backslash \{w_1,w_2\}\nrightarrow \mathcal{V}_1\backslash \{w_1\}.$$ Hence, by (\ref{e39}) and (\ref{e22}), we have $w_1\rightarrow \mathcal{V}_1\backslash \{w_1\}$. Now we have $w_1\rightarrow w_2\rightarrow w$ and $w_1\rightarrow w_0\rightarrow w$, a contradiction.

{\it Case 2.} (\ref{e14}) and (\ref{e15}) hold. Then
\begin{equation}\label{e6}
e(\mathcal{V}_1)=\Delta^+-1.
\end{equation}
(\ref{e6}) implies that there exists a vertex $x\in \mathcal{V}_1$, which has no predecessor in $\mathcal{V}_1$, such that
\begin{equation}\label{eq2}\mathcal{V}_1\rightarrow \mathcal{V}_1\backslash \{x\}
 \end{equation}
Note that $|\mathcal{V}_1|=|\mathcal{V}_2|+1$. By Lemma \ref{le5}, there exists a unique vertex $y\in \mathcal{V}_1$ such that $e(y,\mathcal{V}_2)=0$ and
\begin{equation}\label{e7}\mathcal{V}_1\backslash \{y\}~{\rm matches}~ \mathcal{V}_2.
\end{equation}

Now we distinguish three cases.

{\it Subcase 2.1.} $\mathcal{V}_4$ is empty, which implies that
\begin{equation}\label{e8}N^+(u)=\mathcal{V}_1 ~{\rm ~for~ all~} u\in \mathcal{V}_2.
\end{equation}
Suppose $y\ne x$. Then $y$ has a predecessor in $\mathcal{V}_1$. By Lemma \ref{le3}, $y$ has no successor in $\mathcal{V}_1$. It follows that there exists no 2-walks in $D(\mathcal{V}_1)$. Recalling (\ref{eq2}), among the vertices of $\mathcal{V}_1$ only $x$ has successors in $\mathcal{V}_1$, which implies $D(\mathcal{V}_1)=S(x)$ where $x\in \mathcal{V}_1\backslash \{y\}$. Moreover, by (\ref{e7}) $x$ has a successor in $\mathcal{V}_2$. Combining with (\ref{e7}) and (\ref{e8}), $D$ is an isomorphism of a special case of $D_4$, where $\mathcal{V}_3=\mathcal{V}_2$.

Suppose $y=x$. By Lemma \ref{le3}, all 2-walks originate at $y$. For any $(u_1,u_2)$ in $D(\mathcal{V}_1)$, either $u_1=y$ or $u_1\in N^+(y)$. From (\ref{eq2}) we have $D(\mathcal{V}_1)=T(y)$. Combing with (\ref{e7}) and (\ref{e8}), $D$ is an isomorphism of $D_3$.

{\it Subcase 2.2.} There exists an arc $(u_1, u_2)$ in $D(\mathcal{V}_2)$ with $N^+(u_2)=\mathcal{V}_1$. By Lemma \ref{le6}, we have $N^+_{\mathcal{V}_1}(u_1)\nrightarrow \mathcal{V}_1$. By Lemma \ref{le5}, we have $e(u_1,\mathcal{V}_2)= 1$ and $e(u_1,\mathcal{V}_1)=\frac{n-1}{2}$. Hence, we have $N^+_{\mathcal{V}_1}(u_1)= \mathcal{V}_1\backslash \{u_1'\}$. By (\ref{eq2}), we obtain $u_1'\rightarrow \mathcal{V}_1\backslash \{x\}$. Since $D$ is loopless, we have $u_1'=x$ and
\begin{equation}\label{e11}D(\mathcal{V}_1)=S(x).
\end{equation}
By Lemma \ref{le5} and (\ref{e15}), we obtain
\begin{equation*}e(u,\mathcal{V}_1)\ge \Delta^+-1 {\rm ~for~ all~} u\in \mathcal{V}_2.
\end{equation*}
\indent We assert that \begin{equation}\label{e9}
 N^+_{\mathcal{V}_1}(u)=\mathcal{V}_1\backslash \{x\} ~{\rm ~for~ all}~ u\in \mathcal{V}_4.
  \end{equation}
  Otherwise, there exists an arc $(t_1, t_2)$ in $D(\mathcal{V}_2)$ with $t_1\rightarrow x$. Then $e(x,\mathcal{V}_1)=\frac{n-1}{2}$ and $e(t_2,\mathcal{V}_1)\ge\frac{n-1}{2}$ contradict Lemma \ref{le2}.

If $x\ne y$, then $x$ has a successor $x_1\in \mathcal{V}_2$. We assert that \begin{equation}\label{e10}
N^+(x_1)=\mathcal{V}_1.
  \end{equation} Otherwise, $x_1$ has a successor $x_2\in \mathcal{V}_2$ and $x\rightarrow x_1\rightarrow x_2$. On the other hand, by (\ref{e7}) and $x\rightarrow x_1$ we have $x\rightarrow \mathcal{V}_1\backslash \{x\}\rightarrow x_2$. We get two 2-walks from $x$ to $x_2$, a contradiction.

We also assert that if there exists $u_3\rightarrow u_4\rightarrow u_5$ in $D(\mathcal{V}_2)$, then $u_5=x_1$. Otherwise, from (\ref{e9}) we have $u_3\rightarrow \mathcal{V}_1\backslash \{x\}$. By (\ref{e7}) and $x\rightarrow x_1$, we have $u_3\rightarrow \mathcal{V}_1\backslash \{x\}\rightarrow u_5$, a contradiction.

It follows that all 2-walks in $D(\mathcal{V}_2)$ share the same terminal vertex $x_1$. Hence for any arc $t_1\rightarrow t_2$ in $D(\mathcal{V}_2)$, if $N^+(t_2)\ne \mathcal{V}_1$, then $x_1$ is the successor of $t_2$ in $\mathcal{V}_2$. Therefore, $$D(\mathcal{V}_2)=T'(x_1)+\sum\limits_{i=1}^kS'(u_i)+aC_1,$$ with $\mathcal{V}_3\ne \mathcal{V}_2$. Combining with (\ref{e7}), (\ref{e11}) and (\ref{e9}), $D$ is an isomorphism of $D_4$ with $\mathcal{V}_3\ne \mathcal{V}_2$.

Now suppose $x=y$. Then $x$ has no successor in $\mathcal{V}_2$. Let us turn back to consider $u_1\rightarrow u_2$ with $u_1,u_2\in \mathcal{V}_2$. From (\ref{e9}) we obtain $N^+_{\mathcal{V}_1}(u_1)=\mathcal{V}_1\backslash \{x\}$. Then $\mathcal{V}_2(u_1)=\{x\}\cup \mathcal{V}_2\backslash \{u_2\}$. Moreover, in $\mathcal{V}_2(u_1)$ there exist $\frac{n-3}{2}$ vertices with outdegree $\Delta^+$ and one vertex with outdegree $\Delta^+-1$. Replace the role of $v$ by $u_1$ and apply the same arguments as in Case 1, $D$ is an isomorphism of $D_2$.

{\it Subcase 2.3.} $\mathcal{V}_4$ is a proper subset of $\mathcal{V}_2$ and we have $u_2\in \mathcal{V}_4$ for every arc $(u_1,u_2)$ in $D(\mathcal{V}_2)$. Suppose $t_1\rightarrow t_2$ with $t_1,t_2\in \mathcal{V}_2$ and $e(t_2,\mathcal{V}_1)=\Delta^+-1$. Then $t_2$ has a successor $t_3\in \mathcal{V}_2$. Similarly, $t_3$ has a successor $t_4\in \mathcal{V}_2$. Since $t_1\rightarrow t_2\rightarrow t_3$ and $t_1\rightarrow \mathcal{V}_1\backslash \{t_1'\}$, then $\mathcal{V}_1\backslash \{t_1'\}\nrightarrow t_3$. Combining with (\ref{e7}), we have $t_1'\rightarrow t_3$. Similarly, $t_2'\rightarrow t_4$. From (\ref{e7}) we obtain $t_1'\ne t_2'$. Since $t_1\rightarrow t_2\rightarrow \mathcal{V}_1\backslash \{t_2'\}$ and $t_1\rightarrow \mathcal{V}_1\backslash \{t_1'\}$, we have $\mathcal{V}_1\backslash \{t_1'\}\nrightarrow \mathcal{V}_1\backslash \{t_2'\}$. From (\ref{eq2}), we obtain $t_1'\rightarrow \mathcal{V}_1\backslash \{x,t_2'\}$, which implies $e(t_1',\mathcal{V}_1)\ge \Delta^+-2$. Similarly, $e(t_2',\mathcal{V}_1)\ge \Delta^+-2$. Hence there exists $t\in \mathcal{V}_1$ such that $t_i'\rightarrow t$ for $i=1,2$, which contradicts Lemma \ref{le2}.

{\it Case 3.} (\ref{e13}) and (\ref{e15}) holds, and
\begin{equation}\label{e12}
 e(\mathcal{V}_1,\mathcal{V}_2)=n-\Delta^+-1.
\end{equation}
By Lemma \ref{le5}, there exist $|\mathcal{V}_2|-1$ vertices of $\mathcal{V}_1$ having exactly one successor in $\mathcal{V}_2$. Applying Lemma \ref{le2}, there exists one vertex $x\in \mathcal{V}_2$ with no predecessor in $\mathcal{V}_1$. Note that $|\mathcal{V}_1|=|\mathcal{V}_2|+1$. There exist $y_1,y_2\in \mathcal{V}_1$ such that $e(y_i,\mathcal{V}_2)=0$ for $i=1,2$ and
\begin{equation}\label{e18}
\mathcal{V}_1\backslash \{y_1,y_2\}{\rm~ matches~} \mathcal{V}_2\backslash \{x\}.
\end{equation}

Now we distinguish two cases.

{\it Subcase 3.1.} $\mathcal{V}_4$ is empty. Then (\ref{e8}) holds.
From (\ref{e16}), there exist cycles in $D(\mathcal{V}_1)$. Assume $t_1\in \mathcal{V}_1$ is in a cycle of $D(\mathcal{V}_1)$. Then there exists $t_1\rightarrow t_2\rightarrow t_3$ in $D(\mathcal{V}_1)$. Lemma \ref{le3} guarantees that only $y_1,y_2$ could be in a cycle, which implies there is only one cycle $y_1\leftrightarrow y_2$ in $D(\mathcal{V}_1)$. (\ref{e16}) implies that each vertex in $\mathcal{V}_1$ has a unique predecessor in $\mathcal{V}_1$. By Lemma \ref{le3}, $\mathcal{V}_1\subset N^+(y_1)\cup N^+(y_2)\cup N^+(N^+(y_1)\cup N^+(y_2))$. Hence, we have $D(\mathcal{V}_1)=T(y_1,y_2)$. Combining with (\ref{e8}) and (\ref{e18}), $D$ is an isomorphism of $D_5$.

{\it Subcase 3.2.} $\mathcal{V}_4$ is a proper subset of $\mathcal{V}_2$. Then there exists an arc $u_1\rightarrow u_2$ in $D(\mathcal{V}_2)$. By Lemma \ref{le6}, we have  $u_1'\rightarrow \mathcal{V}_1\backslash \{u_1'\}$.

We assert that
\begin{equation}\label{e20}u\rightarrow \mathcal{V}_1\backslash \{u_1'\} ~{\rm ~for~ all~} u\in \mathcal{V}_4.
\end{equation} Otherwise, there exists $u\in \mathcal{V}_4$ such that $u'\ne u_1'$. Then $u$ has a successor $u^*\in \mathcal{V}_2$. By Lemma \ref{le6}, we obtain $u'\rightarrow \mathcal{V}_1\backslash \{u'\}$, a contradiction with Lemma \ref{le2}.

We assert if there exists $t_1\rightarrow t_2\rightarrow t_3$ in $D(\mathcal{V}_2)$, then either $u_1'\rightarrow t_3$ or $t_3=x$. Otherwise, $t_3$ has a predecessor $t_4\in \mathcal{V}_1\backslash \{u_1'\}$. Since $t_1\in \mathcal{V}_4$ and (\ref{e20}), we have $t_1\rightarrow t_4\rightarrow t_3$, a contradiction.

Let us turn back to consider $u_1\rightarrow u_2$ with $u_1,u_2\in \mathcal{V}_2$. Now by Lemma \ref{le6}, we have $u_2\in \mathcal{V}_4$, which means $u_2$ has a successor in $\mathcal{V}_2$. Applying Lemma \ref{le6} repeatedly, we see that $D(\mathcal{V}_2)$ must contain a cycle $C_s$. Since $u_1'$ has a unique successor $z\in \mathcal{V}_2$, according to the above assertion $D(\mathcal{V}_2)$ contains only one 2-cycle $z\leftrightarrow x$. Moreover, for any arc $(t_1,t_2)$ in $D(\mathcal{V}_2)$, either $t_1\in N^-(x)\cup N^-(z)$ or $t_2\in N^-(x)\cup N^-(z)$. By Lemma \ref{le5}, we have
\begin{equation}\label{e21}
D(\mathcal{V}_2)=T'(x,z)+aC_1,
\end{equation}where $a\ge 1$.

From (\ref{e16}), $u_1'$ has a unique predecessor $u_1^*\in \mathcal{V}_1$. By Claim 1 we have $u_1^*\in \{y_1,y_2\}$. Without loss of generality, we let $y_1\rightarrow u_1'$. Note that $x\in \mathcal{V}_4$ as $x\rightarrow z$. From (\ref{e20}) we get $x'=u_1'$. Hence, $D(\mathcal{V}_1)=S_{y_1}(x')$. Combing with (\ref{e18}), (\ref{e20}) and (\ref{e21}), $D$ is an isomorphism of $D_6$.

So far we have proved that if $e(D)=ex(n)$, then $D$ is an isomorphism of $D_i$ with $i\in \{1,\ldots,6\}$. On the other hand, by Lemma \ref{le8}, $D_i$ is $ \mathscr{F}$-free for $i\in \{1,\ldots,6\}$. Hence, we get the second part of Theorem \ref{th1}.

This completes the proof.

\end{proof}
\section*{Acknowledgement}

Partial of this work was done when Huang was visiting Georgia Institute of Technology and Lyu was visiting Auburn University with the financial support of China Scholarship Council. They thank China Scholarship Council,  Georgia Tech and Auburn University for their support. The second author also thanks Professor Tin-Yau Tam for helpful discussions on matrix theory during his visit.

redpajama_set_name: RedPajamaArXiv